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Electromagnetic Induction - Practice Questions (39)

Question 1: 1. Physicists through hard experiments to explore the physical laws of nature, for human science has...

1. Physicists through hard experiments to explore the physical laws of nature, for human science has made a great contribution, worthy of our admiration. The following description is not consistent with the historical facts of physics ( )

  • A. A. Corrugator discovered Corrugator's Law which determines the direction of induced currents
  • B. B. Lorentz experimentally determined the force of a magnetic field on an electric current
  • C. C. Auster discovered the magnetic effect of electric current
  • D. D. Ampere developed the molecular current hypothesis

Answer: B

Solution: A. Corrugator discovered the law of determining the direction of induced current, namely, Corrugator's law, so A is correct; B. Ampere determined the force of a magnetic field on an electric current by experiment, so B is wrong; C. Auster discovered the existence of magnetic fields around the current, that is, the magnetic effect of the current, so C is correct; D. Ampere put forward the molecular current hypothesis, so D is correct; Not in line with the historical facts of physics is therefore selected B.

Question 2: 2. As shown in the figure, $a b$ is part of a closed circuit in a uniform magnetic field perpendicul...

2. As shown in the figure, $a b$ is part of a closed circuit in a uniform magnetic field perpendicular to the paper surface outward, and the following statements are correct. ![](/images/questions/phys-electromagnetic-induction/image-001.jpg)

  • A. A. When $a b$ moves vertically outward from the paper surface, there is an induced current in $a b$ B. When $a b$ moves vertically inward from the paper surface, there is an induced current in $a b$
  • B. B. When $a b$ is moving vertically outward from the paper surface, there is an induced current in $a b$ B. When $a b$ is moving vertically inward from the paper surface, $a b$
  • C. C. When $a b$ is flattened to the right perpendicular to the line of magnetic inductance, there is an induced current in $a b$
  • D. D. When $a b$ is flattened to the left perpendicular to the line of magnetic inductance, there is no induced current in $a b$

Answer: C

Solution: AB . When part of the conductor of a closed circuit cuts a line of magnetic induction, there must be an induced current in the circuit, so $a b$ does not cut the line of magnetic induction when it moves perpendicularly to the paper surface outward and inward, and there is no induced current in $a b$, so AB is incorrect; CD . When $a b$ moves perpendicular to the magnetic induction to the right and to the left, it cuts the magnetic induction, and there is induced current in $a b$, so C is correct and D is wrong.

Question 3: 3. As shown in the figure, a uniform magnetic field perpendicular to the circular coil pointing to t...

3. As shown in the figure, a uniform magnetic field perpendicular to the circular coil pointing to the paper, $a , b , c , d$ for the circular coil on the four points at equal distances, is now an external force acting on the above four points, the coil will be pulled into a square. If the coil wire is not extendable and the coil is still in the plane where it was originally located, then in the process of deformation of the coil ( ![](/images/questions/phys-electromagnetic-induction/image-002.jpg)

  • A. A. An induced current in the $a d c b$ direction will be generated in the coil.
  • B. B. The induced current in the coil is generated in the direction of $a b c d$ and then $a d c b$.
  • C. C. An induced current in the $a b c d$ direction will be generated in the coil.
  • D. D. No induced current is generated in the coil

Answer: C

Solution: When the magnetic flux becomes smaller when it changes from a circle to a square, an induced current in the direction of abcda will be generated in the coil according to Corrugator's Law. Therefore, choose C.

Question 4: 4. With respect to induced electromotive force, one of the following statements is correct (

4. With respect to induced electromotive force, one of the following statements is correct (

  • A. A. The power supply electromotive force is the induced electromotive force
  • B. B. The part of the conductor that generates the induced electromotive force is equivalent to the power source
  • C. C. In the phenomenon of electromagnetic induction there must be no induced electromotive force without induced current.
  • D. D. If there is current in a circuit, there must be an induced electromotive force

Answer: B

Solution: The source of the electric potential of the power supply is many, not necessarily due to electromagnetic induction, so option A is wrong; in the phenomenon of electromagnetic induction, the part of the conductor that generates the induced electromotive force is equivalent to the power supply, if the circuit is not closed there is no no induction The current in the circuit may be supplied by a chemical cell or other battery as a power source, so there is a current and not necessarily an induced electromotive force, D is wrong.

Question 5: 5. On the phenomenon of electromagnetic induction, the following statements are correct ( )

5. On the phenomenon of electromagnetic induction, the following statements are correct ( )

  • A. A. The greater the magnetic flux through the coil, the greater the induced electromotive force generated in the coil
  • B. B. The greater the change in magnetic flux through the coil, the greater the induced electromotive force generated in the coil
  • C. C. The magnetic flux through the coil is zero and the induced electromotive force generated in the coil must be zero
  • D. D. The faster the flux through the coil changes, the greater the induced electromotive force generated in the coil

Answer: D

Solution: According to Faraday's Law of Electromagnetic Induction, the faster the flux through the coil changes, the greater the induced electromotive force generated inside the coil, and the induced electromotive force and the flux through the coil, through the coil of the flux changes are not directly related to the size of the flux through the coil is zero, the induced electromotive force generated inside the coil is not necessarily zero, so ABC is wrong, D is correct.

Question 6: 6. As shown in the figure, there is a rectangular coil abcd on the right side of a straight wire wit...

6. As shown in the figure, there is a rectangular coil abcd on the right side of a straight wire with constant current, and the wire is coplanar with the coil. If the coil moves with an induced current in the direction of abcda, the possible motion of the coil is ( ) [IMAGE_0]]

  • A. A. pan upwards
  • B. B. pan down
  • C. C. pan left
  • D. D. pan right

Answer: C

Solution: When the current strength in the wire is unchanged, the magnetic field generated is unchanged, and the magnetic induction strength around the wire is unchanged, then the magnetic flux through the wire frame is unchanged, i.e., no induced current will be generated, so AB is incorrect; when the wire frame is moving to the left, the magnetic induction strength in the wire frame increases, and the magnetic flux through the wire frame increases, so an induced current can be generated, and according to Corrugated's Law, the direction of the current can be known as [[INLINE_FORMULA_0 0]], so C is correct; when the wireframe moves to the right, the magnetic induction decreases in the wireframe, the magnetic flux through the wireframe decreases, and an induced current can be generated, according to the law of corrugator, the direction of the current is $a d c b a$, so D is wrong. Therefore, C is correct and ABD is wrong.

Question 7: 7. The scientist who made an important contribution to the establishment of Newton's first law was (...

7. The scientist who made an important contribution to the establishment of Newton's first law was ( ).

  • A. A. Henry Cavendish (1731-1810), English nobleman and pioneer experimental scientist
  • B. B. Michael Faraday (1791-1867), British experimental physicist prominent in the development of electricity
  • C. C. Galileo Galilei (1564-1642), Italian scientist
  • D. D. Oster (brand)

Answer: C

Solution: Before Newton discovered Newton's first law, Galileo studied the relationship between force and motion and concluded that force is not the cause of maintaining the motion of an object, laying the foundation for the establishment of Newton's first law, so C is correct and ABD is wrong.

Question 8: A closed rectangular coil $a b c d$ rotates at uniform speed around a fixed axis $O O ^ { \prime }$ ...

A closed rectangular coil $a b c d$ rotates at uniform speed around a fixed axis $O O ^ { \prime }$ perpendicular to the magnetic susceptibility, and the plane of the coil is located in a uniform magnetic field as shown in Figure 1. The magnetic flux $\Phi$ through the coil changes with time $t$ as shown in Fig. 2, the following statements are correct ( ) ![](/images/questions/phys-electromagnetic-induction/image-004.jpg) A ![](/images/questions/phys-electromagnetic-induction/image-005.jpg) B

  • A. A. $t _ { 1 } , t _ { 3 }$ The rate of change of magnetic flux through the coil is maximum at the time of $t _ { 1 } , t _ { 3 }$.
  • B. B. $t _ { 1 } , t _ { 3 }$ The direction of the induced current in the coil changes at the moment.
  • C. C. $t _ { 2 } , t _ { 4 }$ Moment of maximum magnetic flux in the coil
  • D. D. $t _ { 3 } , t _ { 4 }$ The induced electromotive force in the coil is minimized at the moment of $t _ { 3 } , t _ { 4 }$.

Answer: B

Solution: A. From Fig. 2, it can be seen that the magnetic flux through the coil at the time of $t _ { 1 } , t _ { 3 }$ is the largest, but the slope is 0, that is, the rate of change of magnetic flux is the smallest, so A is wrong; B. The magnetic flux through the coil at the $t _ { 1 } , t _ { 3 }$ time is maximum, the induced current is 0, the coil is in the neutral plane, and the direction of the induced current changes, so B is correct; C. From Fig. 2, we can see that the magnetic flux in the coil is the smallest at the $t _ { 2 } , t _ { 4 }$ time, so C is wrong; D. $t _ { 3 }$ time coil in the magnetic flux is the largest, the rate of change of magnetic flux is the smallest that is the minimum induced electromotive force, but $t _ { 4 }$ time coil in the magnetic flux is the smallest, the rate of change of magnetic flux is the most, that is, the induced electromotive force is the largest, so D is wrong.

Question 9: 9. The magnetic flux through a single-turn closed coil increases uniformly by 12 wb in 6 s. During t...

9. The magnetic flux through a single-turn closed coil increases uniformly by 12 wb in 6 s. During this process, the magnetic flux in a single-turn closed coil increases uniformly by 12 wb.

  • A. A. The induced electromotive force in the coil will increase uniformly
  • B. B. The induced current in the coil will increase uniformly
  • C. C. The induced electromotive force in the coil will remain constant at 2 V.
  • D. D. The induced current in the coil will remain unchanged at 2A.

Answer: C

Solution: By Faraday's law of electromagnetic induction, the induced electromotive force $E = n \frac { \Delta \Phi } { \Delta t } = 1 \times \frac { 12 W b } { 6 s } = 2 V$, the magnetic flux of the closed coil increases continuously and uniformly by 12 Wb for 6 seconds, then the induced electromotive force is a constant value that does not change with time, so A is wrong, C is correct; according to Ohm's law, the induced current in the coil is a constant value, so B is wrong; because the resistance of the coil is not known, then it is impossible to determine the The resistance of the coil is not known, so it is impossible to determine the size of the current, so D is wrong. Therefore, C. The key to solving this problem is to master Faraday's law of electromagnetic induction $E = n \frac { \Delta \Phi } { \Delta t }$, we must know that when the magnetic flux varies uniformly, the induced electromotive force is constant, master the application of Ohm's law, and pay attention to the coil closure to have a current.

Question 10: 10. With respect to the induced electromotive force in a coil, one of the following statements is co...

10. With respect to the induced electromotive force in a coil, one of the following statements is correct

  • A. A. The stronger the coil is placed in the magnetic field, the greater the induced electromotive force in the coil
  • B. B. The greater the magnetic flux through the coil, the greater the induced electromotive force in the coil
  • C. C. The greater the change in magnetic flux through the coil, the greater the induced electromotive force in the coil
  • D. D. The faster the magnetic flux through the coil changes, the greater the induced electromotive force in the coil

Answer: D

Solution: According to Faraday's law of electromagnetic induction, the magnitude of the induced electromotive force is proportional to the rate of change of the magnetic flux through the coil, i.e., the faster the flux changes, the greater the induced electromotive force in the coil.

Question 11: 11. As shown in the figure, insulated metal rings $a , b$ are placed on the left and right sides of ...

11. As shown in the figure, insulated metal rings $a , b$ are placed on the left and right sides of the coil, and at the moment when the conductor bar AB is dropped in a uniform magnetic field, the $a , b$ ring will ![](/images/questions/phys-electromagnetic-induction/image-006.jpg)

  • A. A. Leaning towards the coil
  • B. B. Jump away to the sides.
  • C. C. Move together to the left.
  • D. D. Move together to the right.

Answer: B

Solution: The moment the conductor rod AB falls in a uniform magnetic field, the magnetic field of the energized spiral tube is strengthened, resulting in an increase in the magnetic flux through the ring, according to corrugated law, the two rings in order to impede the increase in magnetic flux to the weak sides of the magnetic field movement, so B is correct, ACD is wrong.

Question 12: 12. The basic structure of an electric pickup in an electric guitar is shown in Fig. The metal strin...

12. The basic structure of an electric pickup in an electric guitar is shown in Fig. The metal strings in the vicinity of the magnet are magnetized and are capable of being magnetized What is usually referred to as a magnetic material is a strongly magnetic substance. Ferromagnetic materials are generally $\mathrm { Fe } , \mathrm { Co } , \mathrm { Ni }$ elements and their alloys, rare earth elements elements and their alloys, rare earth elements and their alloys, and some compounds of Mn. Therefore, when a string vibrates, an induced current is generated in the coil, and the current is amplified by the circuit and transmitted to the loudspeaker to emit sound, the following statements are incorrect ( ) ![](/images/questions/phys-electromagnetic-induction/image-007.jpg)

  • A. A. Remove the magnets and the electric guitar will not function properly.
  • B. B. Increasing the number of coils increases the induced electromotive force in the coil.
  • C. C. The direction of the current in the coil is constantly changing during the vibration of the string
  • D. D. Choose copper strings and the electric guitar will still work fine

Answer: D

Solution: A. If the magnet is taken away, there will be no magnetic field, and when the string vibrates, no induced current will be generated in the coil, and the electric guitar will not work properly, so A does not fit the meaning of the question; B. According to $E = n \frac { \widehat { \lambda } \Phi } { \Delta t }$, increasing the number of turns in the coil can increase the induced electromotive force in the coil; C. string vibration process, the direction of the magnetic field is unchanged, by then the magnetic flux is sometimes larger, sometimes smaller, then the current in the coil C. The direction of the magnetic field remains the same during the vibration of the string; D. Copper can not be magnetized, the choice of copper strings, electric guitars can not work properly, D in line with the meaning of the question.

Question 13: 13. As shown in the figure, there is a constant current of fixed straight wire to the right of a rec...

13. As shown in the figure, there is a constant current of fixed straight wire to the right of a rectangular coil, wire and coil placed in the same smooth horizontal plane, if you reduce the current strength in the wire, the coil will ( ) ![](/images/questions/phys-electromagnetic-induction/image-008.jpg)

  • A. A. pan left
  • B. B. hold still
  • C. C. pan right
  • D. D. swing

Answer: A

Solution: When the current strength in the wire decreases, the magnetic field around the energized wire decreases, through the rectangular coil perpendicular to the paper surface outward magnetic flux decreases, by the law of fluting can be known as a rectangular coil of induced current in the direction of the counterclockwise direction, according to the rule of the left hand of the rectangular coil of the combined Ampere's force is horizontal to the left, so the rectangular coil to the left, so the rectangular coil translates to the left, so the A is correct, B, C, D is wrong;

Question 14: 14. As shown in the figure, the solenoid is connected to a sensitive ammeter, a shaped magnet is rel...

14. As shown in the figure, the solenoid is connected to a sensitive ammeter, a shaped magnet is released from above the solenoid from rest and passes down through the solenoid, regardless of air resistance. The following statements are correct ![](/images/questions/phys-electromagnetic-induction/image-009.jpg) ![](/images/questions/phys-electromagnetic-induction/image-010.jpg) ## Potential below $B$ point after

  • A. A. A bar magnet in free-fall motion
  • B. B. The current in the ammeter goes from $B$ to $A$ and then from $A$ to $B$.
  • C. C. The reduction in gravitational potential energy of the magnet is equal to the heat generated in the loop
  • D. D. The potential at point $A$ is first higher than the potential at point $B$, the

Answer: B

Solution: A. Due to the magnet in motion to produce electromagnetic induction phenomenon, so the magnet is hindered by the role of the magnet, so the magnet will not do free-fall motion; therefore, A error; BD. When the magnet N pole downward movement, resulting in the magnetic flux through the coil becomes larger, and the direction of downward, then by the corrugated law can be produced in the coil of the direction of the induced current circling down, the lower end of the solenoid is equivalent to the positive pole of the power supply. So the direction of current through G is from $B$ to $A$, and when the S-pole is out of the solenoid, the magnetic flux through the coil becomes smaller and the direction is downward, and the upper end of the solenoid is equivalent to the positive pole of the power supply. So the direction of current through G is from $A$ to $B$, then the potential at $A$ is lower than that at $B$, then higher than that at $B$, so the potential at $B$ is higher than that at $B$. The potential at $D$ is wrong and $B$ is correct; C. According to the law of conservation of energy, the gravitational potential energy reduced by the magnet is equal to the sum of the heat generated in the circuit and the kinetic energy of the magnet, so the gravitational potential energy reduced by the magnet is greater than the heat generated in the circuit, so C is wrong.

Question 15: 15. As shown in the figure, the square wire frame $a b c d$ placed in a uniform magnetic field is st...

15. As shown in the figure, the square wire frame $a b c d$ placed in a uniform magnetic field is stationary, the direction of the magnetic field is perpendicular to the plane of the wire frame, the magnetic induction $B$ with the time $t$ of the change of the relationship between the Figure B, $t = 0$ moment, the direction of the magnetic induction perpendicular to the plane of the paper, $t = 0$ time, the magnetic induction direction perpendicular to the plane of the paper. FORMULA_3]] time, the direction of magnetic induction is perpendicular to the paper surface. at . The following options can represent the relationship between the change in the amperometric force $F$ on the $F$ side of the wireframe and the change in the time $t$ (with the horizontal to the left as the positive direction of the $F$). ![](/images/questions/phys-electromagnetic-induction/image-011.jpg) A ![](/images/questions/phys-electromagnetic-induction/image-012.jpg) B

  • A. A. ![](/images/questions/phys-electromagnetic-induction/image-001.jpg)
  • B. B. ![](/images/questions/phys-electromagnetic-induction/image-002.jpg)
  • C. C. ![](/images/questions/phys-electromagnetic-induction/image-003.jpg)
  • D. D. ![](/images/questions/phys-electromagnetic-induction/image-004.jpg)

Answer: A

Solution: $0 \sim 1 \mathrm {~s}$ inside the magnetic field direction inward and uniformly decreasing, by the corrugator's law can be obtained in the coil to produce a clockwise direction of the induced current, by the equation $$ E = \frac { \Delta \Phi } { \Delta t } = \frac { \Delta B } { \Delta t } S $$ It can be seen that the induced electromotive force generated is constant, due to the $$ I = \frac { E } { R } $$ Then the current is constant due to $$ F = B I L $$ It is known that the magnetic induction decreases uniformly, then the $a b$ side of the wireframe is subjected to an amperometric force $F$ that decreases uniformly, and by the left hand rule the amperage The direction of the force is horizontally to the left and is positive. $1 \sim 3 s$ The direction of the internal magnetic field is outward and uniformly increasing, by the corrugated law can be obtained in the coil to produce clockwise induced current, the induced electromotive force is constant, the current is constant, according to the formula of the amperometric force can be known as with the uniform increase in the intensity of the magnetic induction, the amperometric force uniformly increases, by the left hand rule of the direction of the amperometric force can be known to be horizontal to the right, for negative values. $3 \sim 5 s$ The direction of the internal magnetic field is outward and decreases uniformly, by the corrugated law can be produced in the coil counterclockwise direction of the induced current, the induced electromotive force is constant, the current is constant, then according to the formula of the amperometric force can be known as, with the magnetic induction intensity decreases uniformly, the amperometric force decreases uniformly, by the left-hand rule can be known as the direction of the amperometric force is horizontally to the left, is a positive value. $5 \sim 6 \mathrm {~s}$ The direction of the internal magnetic field is inward and uniformly increasing, by the corrugator's law can be obtained from the coil to produce counterclockwise induced current, the induced electromotive force is constant, the current is constant, then according to the formula of the amperometric force can be known as with the magnetic induction strength uniformly increasing, the amperometric force uniformly increasing, by the left-hand rule can be known as the direction of the amperometric force level to the right, is a negative value.

Question 16: 16. As shown in Figure A, in the long straight wire $M N$ placed next to a closed wire frame $a b c ...

16. As shown in Figure A, in the long straight wire $M N$ placed next to a closed wire frame $a b c d$, the long straight wire and the wire frame in the same plane, the long straight wire through the alternating current shown in Figure B. The current along $M N$ is positive. Provide that the direction of the current along $M N$ is positive, the following statements are correct ( ) ![](/images/questions/phys-electromagnetic-induction/image-013.jpg) A ![](/images/questions/phys-electromagnetic-induction/image-014.jpg) B

  • A. A. $0 \sim t _ { 1 }$ time, an induced current along $a b c d$ direction is generated in the wire frame
  • B. B. $t _ { 2 }$ moment of maximum magnetic flux through the wireframe
  • C. C. $0 \sim t _ { 1 }$ time, the wireframe is subjected to an amperometric force to the right
  • D. D. $t _ { 1 } \sim t _ { 3 }$ The direction of the amperometric force on the wireframe remains the same for the duration of time

Answer: C

Solution: A. In $0 \sim t _ { 1 }$ time, the current from $M$ to $N$ increases, the magnetic flux through the rectangular wire frame $a b c d$ increases inwardly, and according to Corrugated's law, $a d c b$ direction of induced current is generated in the wire frame. INLINE_FORMULA_4]] direction, so A is wrong; B. The current from $M$ to $N$ at the time of $t _ { 2 }$ is 0, and the magnetic flux through the wire frame is 0, so B is wrong; C. $0 \sim t _ { 1 }$ time, from the analysis of the A option, through the rectangular wire frame $a b c d$ of the magnetic flux inward, the wire frame produced $a d c b$ direction of the induced current, by the left hand rule can be known, the long straight wire is subjected to the right of the amperage force, so C is correct; D. In $t _ { 1 } \sim t _ { 2 }$ time, the current from $M$ to $N$ decreases, the magnetic flux through the rectangular wire frame $a b c d$ decreases inwardly, and according to Corrugator's law, there is a $a b c d$ in the wire frame. $a b c d$ direction of the induced current, by the left hand rule can be seen, the long straight wire by the amperage force to the left; $t _ { 2 } \sim t _ { 3 }$ time, from $N$ to $M$ the current increases, the rectangular wire frame $M$, the rectangular wire frame $M$ through the magnetic flux decreases inward, according to the flute law can be seen, the wire frame produced The magnetic flux through the rectangular wire frame $a b c d$ increases outward, and according to corrugator's law, an induced current in the $a b c d$ direction is generated in the wire frame, and by the left-hand rule, the amperage is directed to the right of the long straight wire; therefore, D is wrong.

Question 17: 17. There are two rings of conductors $a , b$ in close proximity to each other on an insulated smoot...

17. There are two rings of conductors $a , b$ in close proximity to each other on an insulated smooth horizontal plane as shown in the figure. Moving a bar magnet down along their proper center (without reaching that plane), how will $a , b$ move ( ) ![](/images/questions/phys-electromagnetic-induction/image-015.jpg)

  • A. A. $a , b$ will move away from each other
  • B. B. $a , b$ will be close to each other
  • C. C. $a , b$ will not be moved.
  • D. D. impossible judge

Answer: A

Solution: According to $\Phi = B S$, $B$ increases as the magnet moves downward, so the magnetic flux through each ring increases, and an induced current in the same direction is generated in $a , b$, and there is a repulsive force between the two rings, so $a , b$ will move away from each other. _3]] will move away from each other, A is correct.

Question 18: 18. As shown in the figure, $a , b , c$ is a coil in the same plane, in which $a , b$ is a concentri...

18. As shown in the figure, $a , b , c$ is a coil in the same plane, in which $a , b$ is a concentric circle, now give $a$ in the clockwise direction of the current, in the $a$ in the current is gradually increased in the process, $b , c$ in the direction of the induced current ( ). During the process, the direction of the induced current generated in $b , c$ is ( ) ![](/images/questions/phys-electromagnetic-induction/image-016.jpg)

  • A. A. Clockwise in $b , c$.
  • B. B. b. Counterclockwise in $c$.
  • C. C. $b$ is counterclockwise, $c$ is clockwise
  • D. D. $b$ is clockwise, $c$ is counterclockwise

Answer: C

Solution: Firstly, from the meaning of the question, we can see that the current in the coil $a$ is gradually increasing, and the magnetic field generated by the current is also increasing. Secondly, according to the right-hand spiral rule, the current in coil $a$ is in the direction of the magnetic field generated by coil $b$, which is perpendicular to the inward side of the paper, and then according to Corrugated's law, we can tell that the induced current generated by coil $b$ is in the counterclockwise direction, which corresponds to the perpendicular to the magnetic field. The magnetic field is perpendicular to the paper surface. Similarly, according to the right-handed helix rule, the current in coil $a$ is in the direction of the magnetic field generated by coil $c$, which is perpendicular to the outward face of the paper, then according to the corrugated law, we can determine that the induced current in coil $c$ is in the clockwise direction, which corresponds to the generated magnetic field The magnetic field is perpendicular to the inward face of the paper. Therefore, choose C.

Question 19: 19. As shown in the figure, the $P$ ring is located at the mid-plumb plane of the solenoid, when the...

19. As shown in the figure, the $P$ ring is located at the mid-plumb plane of the solenoid, when the induced current in the $P$ ring is clockwise (right-viewing) and subjected to the outward tension away from the center of the circle, then the direction of the current and the change in size of the solenoid is ( ) ![](/images/questions/phys-electromagnetic-induction/image-017.jpg)

  • A. A. $b \rightarrow a$, decrease
  • B. B. $b \rightarrow a$, increase
  • C. C. $a \rightarrow b$, decrease
  • D. D. $a \rightarrow b$, increase

Answer: D

Solution: The induced current in the $P$ ring is clockwise, and according to the right-hand rule, the $P$ ring produces an induced magnetic field that is horizontally to the left, and because the $P$ ring has a tendency to expand, which means that the magnetic field produced by the solenoid is enhanced, then the current through the solenoid is increasing. The current through the solenoid is increasing, and the magnetic field generated by the solenoid is horizontally to the right according to Corrugator's law, and the direction of the current through the solenoid is $a \rightarrow b$ according to the right-hand rule.

Question 20: 20. As shown in the figure, a closed coil is suspended from the ceiling by a thin insulated wire, an...

20. As shown in the figure, a closed coil is suspended from the ceiling by a thin insulated wire, and the closed coil is at rest. A bar magnet is to the right of the metal ring, the bar magnet and the metal ring axis coincide, the bar magnet to the left near the metal ring process $\_\_\_\_$) ![](/images/questions/phys-electromagnetic-induction/image-018.jpg) ![](/images/questions/phys-electromagnetic-induction/image-019.jpg)

  • A. A. The metal ring swings to the right while expanding along the radius
  • B. B. The metal ring swings to the left while expanding along its radius
  • C. C. The metal ring swings to the right while contracting in the direction of the radius
  • D. D. The metal ring swings to the left while contracting in the direction of the radius

Answer: D

Solution: Bar magnet to the left near the metal ring process, through the metal ring of magnetic flux increases, according to corrugated law, the direction of the magnetic field of the induced current and the direction of the magnetic field of the original current is the opposite direction, so mutual repulsion, then the metal ring will be swinging to the left ("come to refuse to go to stay"), due to the flux increases, the metal ring has a tendency to contract ( Increase shrinkage and decrease expansion"), so D is correct, ABC is wrong.

Question 21: 21. As shown in the figure, a closed triangular wire frame $A B C$ is located in a straight plane, b...

21. As shown in the figure, a closed triangular wire frame $A B C$ is located in a straight plane, below which (slightly in front of) is fixed a horizontal straight wire parallel to the plane of the wire frame, in which a constant current in the direction shown is passed. The wire frame is released, and it falls from the solid position to the dashed position without rotation, during which the ![](/images/questions/phys-electromagnetic-induction/image-020.jpg) ![](/images/questions/phys-electromagnetic-induction/image-021.jpg)

  • A. A. The direction of the induced current in the wireframe is clockwise $\rightarrow$ counterclockwise in that order
  • B. B. The induced current is not zero when the magnetic flux of the wire frame is zero
  • C. C. The direction of the combined amperometric force on the wireframe is up $\rightarrow$ down $\rightarrow$ up in that order
  • D. D. The wireframe is subjected to zero combined amperometric force and is in free-fall motion

Answer: B

Solution: A. According to the right-hand rule, the magnetic field of an energized straight wire is outward at the top and inward at the bottom; close to the wire The magnetic field is strong near the wire and weak far away from the wire. During the process of approaching the wire from the top to the bottom of the wire frame, the outward magnetic induction intensity increases, and according to the corrugated law, a clockwise current is generated in the wire frame; when crossing the wire, the upward outward magnetic field and the downward inward magnetic field are superimposed, and the outward magnetic flux decreases first, and then it becomes inward magnetic flux, and it gradually increases until it is the maximum; according to the corrugated law, a counterclockwise current is generated in the wire frame. After the inward magnetic flux becomes maximum, it continues to move downward, and the inward magnetic flux gradually decreases, at which time the new direction of the current becomes clockwise again, i.e., the direction of the induced current is $A C B A \rightarrow A B C A \rightarrow A C B A$ in order. Therefore $A$ is wrong; B. According to the analysis in $A$, when crossing the wire, the magnetic field outward from the top and inward from the bottom are superimposed, and the magnetic flux outward first decreases, and decreases to 0, and then it becomes inward, and gradually increases. This process is continuous and there is always an induced current, so $B$ is correct; $C D$. According to Corrugator's law, the induced current always prevents the motion of the wire frame relative to the magnetic field, so the direction of the amperometric force is always upward, not 0. Therefore, the wire frame cannot be in free fall, so $C$ is wrong, $D$ is wrong.

Question 22: 23. aircraft carrier aircraft ejection smooth insulated rod takeoff utilized by the electromagnetic ...

23. aircraft carrier aircraft ejection smooth insulated rod takeoff utilized by the electromagnetic drive around, the principle as shown in the figure, when the fixed ring coil on the sudden passage of DC current when the coil on the left side of the metal ring was ejected, then the following statement is correct ( ) ![](/images/questions/phys-electromagnetic-induction/image-022.jpg)

  • A. A. At the moment of closing the switch S, an induced current in the clockwise direction is generated in the ring as seen from the right side.
  • B. B. The metal ring will tend to expand as it moves to the left
  • C. C. If the metal ring is placed on the right side of the coil the ring will not be ejected
  • D. D. If the positive and negative terminals of the battery are switched, the metal ring cannot be ejected to the left

Answer: A

Solution: A. The coil of current for the left into the internal magnetic field generated by the direction of the right in the process of closing the switch, the magnetic field becomes stronger, then by the corrugated law can be seen by the metal ring of the induction loop current from the right side of the clockwise. Therefore, option A is consistent with the question. B. In the process of closing the switch, the magnetic field becomes stronger, then by the corrugator's law, the area of the metal ring has a tendency to shrink, so the Therefore, option B does not fit the question. C. If the metal ring is placed on the right side of the coil, the ring will move to the right according to Corrugator's law. Therefore, option C does not fit the question. D. After the positive and negative poles of the battery are switched, according to Corrugated's law, the force on the metal ring is still to the left, so it will still be ejected to the left. Therefore, option D does not fit the question. D does not fit the question.

Question 23: 24. As shown in the figure, a square closed metal wire frame from a certain height free fall, the mi...

24. As shown in the figure, a square closed metal wire frame from a certain height free fall, the middle through a uniform magnetic field, the magnetic field direction perpendicular to the plane of the wire frame. From the beginning of the magnetic field to completely leave the magnetic field, the wire frame in the induced current with time may be $\square$ $\begin{array} { c c c } \times & \times & \times \\ \times & \times & \times \end{array}$

  • A. A. The magnitude and direction remain the same
  • B. B. The size changes while the direction remains the same
  • C. C. The current decreases as it starts to enter the magnetic field and increases as it leaves the magnetic field
  • D. D. The current increases as it starts to enter the magnetic field and decreases as it leaves the field

Answer: D

Solution: AB. When the coil enters the magnetic field, the inward magnetic flux increases; when it leaves the field, the inward magnetic flux decreases, then the direction of the induced current must be opposite, option AB is wrong; CD. If the coil enters the magnetic field in such a way that [BLOCK_FORMULA_0]] i.e., the coil enters the magnetic field with uniform speed when $v _ { 0 } = \frac { m g R } { B ^ { 2 } L ^ { 2 } }$; If the coil enters the magnetic field with a speed of ${ } ^ { v > v _ { 0 } }$, the coil will decelerate into the magnetic field, and the current will decrease during the process; when the coil leaves the magnetic field with a speed that may be equal to $v _ { 0 }$ or greater than $v _ { 0 }$, then the coil may have a uniform speed when it exits the magnetic field, or it may decelerate. The induced current may remain constant or decrease, so option C is wrong; If the coil enters the magnetic field with the speed ${ } ^ { v < v _ { 0 } }$, then the coil will accelerate into the magnetic field, and the current will increase during the process; when the coil leaves the magnetic field with the speed equal to $v _ { 0 }$ or greater than $v _ { 0 }$, then the coil may be uniformly speeding up or decelerating as it exits the magnetic field. The induced current may be constant or decreasing, option D is correct.

Question 24: 25. The following statements about the phenomenon of electromagnetic induction are correct

25. The following statements about the phenomenon of electromagnetic induction are correct

  • A. A. The magnetic flux through the closed circuit increases, but the induced current in the closed circuit may decrease
  • B. B. There can be no induced current in a closed circuit at the instant when the magnetic flux through the closed circuit is zero
  • C. C. If the magnetic flux through the closed circuit decreases, the induced electromotive force in the closed circuit must decrease
  • D. D. The magnetic flux through the closed circuit is changing faster and faster, but the induced electromotive force in the closed circuit may be unchanged

Answer: A

Solution: A, according to Faraday's law of electromagnetic induction is known: the size of the induced electromotive force and the rate of change of magnetic flux is proportional to the magnetic flux has no direct relationship; through the closed circuit of the magnetic flux increases, but the rate of change of magnetic flux does not necessarily increase, if the rate of change decreases, then according to Faraday's law of electromagnetic induction is known: the induced electromotive force decreases, the induced current will be reduced, so A correct. B, through the closed circuit in the magnetic flux is not zero, if the magnetic flux changes, the closed circuit will have an induced current, so B error. C, through the closed circuit in the magnetic flux is reduced, the rate of change of magnetic flux is not necessarily reduced, according to Faraday's fixed electromagnetic induction. According to Faraday's law of electromagnetic induction, the induced electromotive force does not necessarily decrease, and the induced current does not necessarily decrease. Therefore, C is wrong. D, according to Faraday's law of electromagnetic induction: the size of the induced electromotive force and the rate of change of magnetic flux is proportional to the rate of change of magnetic flux, so the faster the change of magnetic flux, the greater the induced electromotive force, the greater the induced current, so D error.

Question 25: 26. As shown in Figure A, the coil $a b$ has a current in it as shown in Figure B. The positive dire...

26. As shown in Figure A, the coil $a b$ has a current in it as shown in Figure B. The positive direction of the current is from $a$ to $b$, and during the time in $0 { } ^ { t _ { 2 } }$, an aluminum ring $M$ suspended by wire generates an induced current. INLINE_FORMULA_4]], an induced current is generated in $M$, then the following statement is correct ![](/images/questions/phys-electromagnetic-induction/image-023.jpg) A ![](/images/questions/phys-electromagnetic-induction/image-024.jpg) B

  • A. A. $0 \sim ^ { t _ { 2 } }$ The magnetic flux through the aluminum ring has been decreasing during this period of time
  • B. B. Looking from left to right, the direction of the induced current is always clockwise
  • C. C. ${ } ^ { t _ { 1 } }$ No inductive current in the aluminum ring.
  • D. D. Looking from left to right, the direction of the induced current is always counterclockwise

Answer: B

Solution: M is in the magnetic field of an energized coil, and the change in magnetic flux depends on the change in current in the coil: $0 \sim ^ { t _ { 1 } }$ During this time the current in the coil decreases, and the magnetic flux through the aluminum ring is decreasing; ${ } ^ { t _ { 1 } } \sim { } ^ { t _ { 2 } }$ During this time the current in the coil is reversed and increased, and the magnetic flux through the aluminum ring is increasing; ${ } ^ { t _ { 1 } }$ During this time the current in the coil is in a state of change, the magnetic flux in the aluminum ring is in a state of change, and there is induced current. AC. INLINE_FORMULA_2]] during which the current in the coil is in a state of variation, the magnetic flux in the aluminum ring is in a state of variation, and there is an induced current.AC is incorrect; BD . According to the meaning of the question, as the current flows from $a$ to $b$ in the positive direction, when the current flows from $a$ to $b$, the direction of the magnetic field of the solenoid is horizontally to the right by the right-handed helix rule, then the magnetic field passing through the aluminum ring is horizontally to the right, and the magnetic field passing through the aluminum ring is horizontally to the right. Then the magnetic field through the aluminum ring flows horizontally to the right, because the current decreases, the magnetic flux becomes smaller, according to corrugator's law, the induced current on the aluminum ring is clockwise (from left to right); when the current is flowing from $b$ to $a$, the right-handed helix rule shows that the magnetic field through the aluminum ring flows horizontally to the left, and the magnetic flux becomes larger when the current is increases, then the magnetic flux becomes larger, according to corrugated law can be obtained, so the induced current clockwise (from left to right), so the direction of the current remains unchanged always clockwise, B is correct, D is wrong.

Question 26: 27. As shown in Fig. A, two closed circular coils A and B, whose centers coincide, are placed in the...

27. As shown in Fig. A, two closed circular coils A and B, whose centers coincide, are placed in the same horizontal plane, and an alternating current is passed through coil B as shown in Fig. B. Let the current be in the clockwise direction at $\mathrm { t } = 0$ (the arrow in the figure shows). For coil A, in $t _ { 1 } \sim t _ { 2 }$ time, the following statements are true ![](/images/questions/phys-electromagnetic-induction/image-025.jpg) A ![](/images/questions/phys-electromagnetic-induction/image-026.jpg) B

  • A. A. There is a clockwise current with a tendency to shrink
  • B. B. There is a clockwise current with a tendency to expand
  • C. C. There is a counterclockwise current with a tendency to shrink
  • D. D. There is a counterclockwise current with a tendency to expand

Answer: A

Solution: The current in coil B produces a magnetic field. In $\mathrm { t } _ { 1 } \sim \mathrm { t } _ { 2 }$ time, when the current in B increases in the reverse direction, according to Ampere's law the direction of the magnetic field in the coil is outward and increases with the increase in the current, so according to Corrugated's law, the magnetic field of the induced current in A is inward, and the direction of the induced current is in the clockwise direction. According to the left-hand rule, this current is subjected to an amperometric force pointing to the center of the circle in the outward magnetic field formed by B, so the coil A tends to shrink.A is correct, and BCD is wrong.

Question 27: 28. As shown in Figure A, the energized solenoid A and the center axis of the coil B suspended by an...

28. As shown in Figure A, the energized solenoid A and the center axis of the coil B suspended by an insulated rope in the same horizontal straight line, A has a change in current as shown in the figure, $t = 0$ when the direction of the current as shown by the arrow in Figure B. In $t _ { 1 } \sim t _ { 2 }$ time, for the direction of current (from left to right) and the direction of motion of coil B, the following judgment is correct ( ) ![](/images/questions/phys-electromagnetic-induction/image-027.jpg) A ![](/images/questions/phys-electromagnetic-induction/image-028.jpg) B

  • A. A. There is a clockwise current in coil B. The coil swings to the right.
  • B. B. There is a clockwise current in coil B. The coil swings to the left.
  • C. C. A counterclockwise current flows in coil B. The coil swings to the right.
  • D. D. A counterclockwise current flows in coil B. The coil swings to the left.

Answer: A

Solution: According to the question, the solenoid is energized in the solenoid, and in ${ } ^ { t _ { 1 } \sim t _ { 2 } }$ time, by the right-handed helix rule, the coil B The magnetic field is horizontal to the left, when the current increases, the magnetic flux becomes larger, according to corrugated law can be obtained, so the induced current clockwise (from the left to the right); from the point of view of the obstruction of the induced current can be known, the coil to the right in order to hinder the increase in magnetic flux.

Question 28: 29. As shown in the figure, in two parallel long straight wire $M , N$, into the same direction and ...

29. As shown in the figure, in two parallel long straight wire $M , N$, into the same direction and size of the current, wire frame $a b c d$ and two wires in the same plane, wire frame along the perpendicular to the direction of the two wires, from right to left in the two wires between the uniform speed, in the process of moving the wire frame in the direction of induced current is () ![](/images/questions/phys-electromagnetic-induction/image-029.jpg)

  • A. A. Along $a b c d a$ No change.
  • B. B. Along adcba No change
  • C. C. Change from abcda to $a d c b a$
  • D. D. Changed from adcba to $a b c d a$

Answer: B

Solution:

Question 29: 30. In a uniform magnetic field, there is a single-turn wire loop connected to a capacitor as shown ...

30. In a uniform magnetic field, there is a single-turn wire loop connected to a capacitor as shown in the figure, and it is known that $C = 30 \mu \mathrm {~F}$ , $l _ { 1 } = 5 \mathrm {~cm} , ~ l _ { 2 } = 8 \mathrm {~cm}$ , and that the magnetic induction increases at a rate of change of ${ } ^ { 5 \times 10 ^ { - 2 } \mathrm {~T} / \mathrm { s } }$, then () ![](/images/questions/phys-electromagnetic-induction/image-030.jpg)

  • A. A. The upper pole plate of the capacitor is positively charged with $6 \times 10 ^ { - 5 } \mathrm { C }$
  • B. B. The upper pole plate of the capacitor is negatively charged with $6 \times 10 ^ { - 5 } \mathrm { C }$
  • C. C. The upper pole plate of the capacitor is positively charged with $6 \times 10 ^ { - 9 } \mathrm { C }$
  • D. D. The upper pole plate of the capacitor is negatively charged with $6 \times 10 ^ { - 9 } \mathrm { C }$

Answer: C

Solution:

Question 30: 31. Coil M and coil P are wound around the same core as shown in the figure. It is known that when c...

31. Coil M and coil P are wound around the same core as shown in the figure. It is known that when current flows in from the left side of the ammeter, the pointer is deflected from the center to the left, then ( ) ![](/images/questions/phys-electromagnetic-induction/image-031.jpg)

  • A. A. At the moment of closing the switch S, the ammeter pointer does not deflect.
  • B. B. At the moment of closing the switch S, the ammeter pointer deviates to the left.
  • C. C. At the moment of disconnecting the switch S, the ammeter pointer does not deflect.
  • D. D. At the moment of disconnecting the switch S, the ammeter pointer deviates to the left.

Answer: D

Solution: AB. At the moment of closing the switch S, the current through the coil M becomes larger, then the magnetic flux through the coil P increases to the left, according to the corrugated law, the induced current in the coil P flows in from the right end of the ammeter, then the pointer will be deflected to the right, the option AB is wrong; CD. At the moment of disconnecting the switch S, the current through the coil M becomes smaller, then the magnetic flux through the coil P decreases to the left, according to the law of corrugator, the induced current in the coil P flows from the left end of the ammeter, then the pointer is deflected to the left, the option C is wrong, and the D is correct.

Question 31: 32. As shown in the figure, straight placed solenoid and wire $a d c d$ constitute a closed circuit,...

32. As shown in the figure, straight placed solenoid and wire $a d c d$ constitute a closed circuit, the circuit around the region of the direction of the perpendicular to the paper in the direction of a uniformly strong magnetic field, solenoids below the level of the desktop there is a conductor ring. To make the conductor ring is subjected to an upward magnetic force, the law of change of magnetic induction with time should be ( ) ![](/images/questions/phys-electromagnetic-induction/image-032.jpg)

  • A. A. ![](/images/questions/phys-electromagnetic-induction/image-005.jpg)
  • B. B. ![](/images/questions/phys-electromagnetic-induction/image-006.jpg)
  • C. C. ![](/images/questions/phys-electromagnetic-induction/image-007.jpg)
  • D. D. ![](/images/questions/phys-electromagnetic-induction/image-008.jpg)

Answer: A

Solution: The ring is interacting with the solenoid, the ring is subjected to an upward force, the magnetic field generated in the solenoid is to become smaller, so the current in the solenoid is to be reduced, and the induced electromotive force is to be reduced under the condition that the resistance remains constant, and the induced electromotive force is proportional to the rate of change of the magnetic flux in the $a b c d$ loop, and the rate of change of the magnetic induction intensity is to be reduced when the area remains constant, i.e., the slope is to be reduced in the [ [INLINE_FORMULA_1]] graph, the slope is to be reduced, i.e.: in the $B - t$ graph, the slope is not only to be changed but also to become smaller, so A is correct and BCD is wrong .

Question 32: 33. A magnetic field magnetic inductance as shown in the figure, there are copper coils from the ill...

33. A magnetic field magnetic inductance as shown in the figure, there are copper coils from the illustrated $A$ position to $B$ position, in the process of falling, from the top down, the direction of the induced current in the coils are ![](/images/questions/phys-electromagnetic-induction/image-033.jpg)

  • A. A. Always clockwise
  • B. B. Always counterclockwise
  • C. C. Clockwise then counterclockwise
  • D. D. Counterclockwise then clockwise

Answer: C

Solution: During the falling process, the magnetic induction intensity increases first and then decreases, so the magnetic flux through the coil increases first and then decreases, from $A$ to $O$, the magnetic flux through the coil increases, so the direction of the magnetic field of the induced current goes downward, the direction of the induced current is clockwise, and from to $B$, the magnetic flux through the coil decreases, so the direction of the magnetic field of the induced current is upward, and the direction of the induced current is counterclockwise.

Question 33: 34. The figure shows the schematic diagram of the principle of wireless charging of a brand of cell ...

34. The figure shows the schematic diagram of the principle of wireless charging of a brand of cell phone. If the magnetic field $B$ generated by the feeding coil is gradually enhanced during a certain period of time, then ![](/images/questions/phys-electromagnetic-induction/image-034.jpg)

  • A. A. At this time, the current in the feeder coil flows through port 1
  • B. B. Port 3 of the coil is positive.
  • C. C. At this point the area of the charged coil tends to expand
  • D. D. The direction of the magnetic field produced by the current in the charged coil is the same as the direction of the magnetic field shown in the diagram

Answer: A

Solution: A. The magnetic field B generated by the feeder coil is gradually increasing during a certain period of time, and according to Ampere's rule, the current in the feeder coil is flowing through port 1 at this time, so A is correct; B. According to corrugator's law, port 4 of the receiving coil is positive, so B is wrong; C. The magnetic field B is gradually increasing, so the area of the charged coil tends to shrink, so C is wrong; D. According to Corrugator's law, the direction of the magnetic field generated by the current in the charged coil is opposite to the direction of the magnetic field shown in the figure, so D is wrong. D. The direction of the magnetic field generated by the current in the charged coil is opposite to that shown in the diagram.

Question 34: 35. Now people can use wireless charging board for cell phone charging, as shown in the figure for t...

35. Now people can use wireless charging board for cell phone charging, as shown in the figure for the charging principle road map, charging board connected to the AC power supply, power supply to the charging board, charging board within the power supply coil can produce alternating magnetic field, so that the cell phone within the receiving coil to produce alternating current, and then by the rectifier circuit into DC power after charging the cell phone battery. If the magnetic field passes upward through the coil perpendicular to the plane of the receiving coil during a certain period of time, its magnetic induction increases uniformly. The following statements are correct. ![](/images/questions/phys-electromagnetic-induction/image-035.jpg)

  • A. A. All cell phones can be charged wirelessly with this wireless charging pad
  • B. B. When charging a cell phone wirelessly, the coil on the phone works on the principle of the "magnetic effect of electric current".
  • C. C. During the above time period, the direction of the induced current is determined by the $c \rightarrow$ receiving coil $\rightarrow d$
  • D. D. During the above time period, the direction of the induced current is determined by the $d \rightarrow$ receiving coil $\rightarrow C$

Answer: C

Solution: A.The cell phone must have a receiving coil and a rectifier circuit for wireless charging.A error; B.When charging wirelessly, the operating principle of the receiving coil on the cell phone is electromagnetic induction.B error ; CD.During the above time period, according to the corrugated law, the direction of the induced current is determined by $c \rightarrow$ the receiving coil $\rightarrow d$.C is correct, D is wrong. C is correct and D is wrong.

Question 35: 36. As shown in the figure, on a smooth insulated horizontal surface, there are two fixed mutually p...

36. As shown in the figure, on a smooth insulated horizontal surface, there are two fixed mutually perpendicular to each other insulated long straight wires, pass the same size of current. In the angular bisector, symmetrically placed four identical circular coils. If the current on the two wires increases uniformly at the same time and according to the same law, the following judgments are correct. ![](/images/questions/phys-electromagnetic-induction/image-036.jpg)

  • A. A. Clockwise induced current in coil 1
  • B. B. There is a clockwise induced current in coil 2
  • C. C. Clockwise induced current in coil 3
  • D. D. There is a clockwise induced current in coil 4

Answer: B

Solution: By Ampere's law, we can determine that the magnetic flux in coil 1 is zero, and the magnetic flux in coil 2 is perpendicular to the horizontal plane outward, and the magnetic flux in coil 3 is zero, and the magnetic flux in coil 4 is perpendicular to the horizontal plane inward. When the currents in the two wires increase at the same time according to the same law, according to Corrugator's law, there is no induced current in coil 1 and coil 3, but there is a clockwise current in coil 2 and a counterclockwise current in coil 4.

Question 36: 37. As shown in Figure A, $a b c d$ is a square, closed metal wireframe in a straight plane with mas...

37. As shown in Figure A, $a b c d$ is a square, closed metal wireframe in a straight plane with mass $m$ and resistance $R$. Below the wire frame there is a uniform magnetic field, $M N$ and $P Q$ are the horizontal boundaries of the uniform magnetic field region and are parallel to the $b c$ side of the wire frame, and the direction of the magnetic field is perpendicular to the plane of the wire frame and inward. Now make the metal wire frame from $M N$ above a certain height from rest to start falling, as shown in Figure B is the metal wire frame from the beginning of the fall to $b c$ just movement to the uniform magnetic field $P Q$ boundary of the $v - t$ graphs The data in the graph are known quantities. The acceleration due to gravity is $g$, regardless of air resistance. The following statement is correct ![](/images/questions/phys-electromagnetic-induction/image-037.jpg) A ![](/images/questions/phys-electromagnetic-induction/image-038.jpg) B

  • A. A. When the metal wire frame first enters the magnetic field, the direction of the induced current is along the $a d c b a$ direction.
  • B. B. The distance between $M N$ and $P Q$ is $v _ { 1 } \left( t _ { 2 } - t _ { 1 } \right)$.
  • C. C. The magnetic induction of the field is $\frac { 1 } { v _ { 1 } \left( t _ { 2 } - t _ { 1 } \right) } \sqrt { \frac { m g R } { v _ { 1 } } }$
  • D. D. The heat generated by the metal wireframe in the time $0 - t _ { 3 }$ is $m g v _ { 1 } \left( t _ { 3 } - t _ { 1 } \right)$

Answer: C

Solution:

Question 37: 38. As shown in the figure, there is a uniform magnetic field in space, has been a metal rod and the...

38. As shown in the figure, there is a uniform magnetic field in space, has been a metal rod and the direction of magnetic induction perpendicular to the direction of the intensity of magnetic induction, when it is at a speed of $v$ along the direction of movement perpendicular to the rod and the intensity of the magnetic induction, the size of the induced electromotive force at the two ends of the rod $\varepsilon$, the rod will be bent into two lengths of equal length and perpendicular to each other, the bend, and the magnetic induction intensity of the two segments. Placed in the plane perpendicular to the magnetic induction intensity, when it along the two sections of the bending line angle bisector When it is moving with speed $v$ in the direction of $v$, the induced electromotive force at both ends of the bar is $\varepsilon ^ { \prime }$, and $\frac { \varepsilon ^ { \prime } } { \varepsilon }$ is equal to ( ). | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | | :---: | :---: | :---: | :---: | :---: | :---: | | $\times$ | $\underset { \times } { \times } v$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\rightarrow _ { x } ^ { v }$ | $\times$ | $\rightarrow _ { x } ^ { v }$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$ | $\times$]

  • A. A. $1 / 2$
  • B. B. $\frac { \sqrt { 2 } } { 2 }$
  • C. C. 1
  • D. D. $\sqrt { 2 }$

Answer: B

Solution: Let the length of the conductor cutting the magnetic field before bending be $L$, and after bending, the effective length of the conductor cutting the magnetic field be $$ l = \sqrt { \left( \frac { L } { 2 } \right) ^ { 2 } + \left( \frac { L } { 2 } \right) ^ { 2 } } = \frac { \sqrt { 2 } } { 2 } L $$ Therefore, the induced electromotive force generated is $$ \varepsilon ^ { \prime } = B l v = B \cdot \frac { \sqrt { 2 } } { 2 } L v = \frac { \sqrt { 2 } } { 2 } \varepsilon $$ So $\frac { \varepsilon ^ { \prime } } { \varepsilon } = \frac { \sqrt { 2 } } { 2 }$ , so ACD is wrong, B is correct.

Question 38: 39. As shown in the figure, in a horizontal table placed on the closed conductor ring is not far abo...

39. As shown in the figure, in a horizontal table placed on the closed conductor ring is not far above, there is a shape of the magnet (S pole up, $N$ pole down) from rest to start moving quickly to the right, about the direction of the induced current in the ring (from the top down) and the interaction of the situation, the following statements are correct ( ) ![](/images/questions/phys-electromagnetic-induction/image-039.jpg)

  • A. A. The direction of the induced current in the ring is counterclockwise
  • B. B. The direction of the induced current in the ring is clockwise
  • C. C. The ring has a tendency to move to the left
  • D. D. The pressure of the ring on the horizontal table increases

Answer: B

Solution: AB . From the diagram, we can see that the direction of the magnetic field in the ring is downward, and when the magnet moves to the right, the magnetic flux through the ring becomes smaller, and from the law of fluting, we can see that the magnetic field generated by the induced current is downward, and then from the rule of Ampere's law, we can see that the induced current in the ring is clockwise when viewed from the top, so A is incorrect, and B is correct; C. By the left-hand rule, the amperometric force on the ring is directed to the upper right, so the ring tends to move to the right, C is wrong; D. In the solid direction of the ring by gravity, desktop support for it, in the process of moving to the right increased Ampere's force along the solid direction of the force, so the support becomes smaller, according to Newton's third law, can be known, the pressure becomes smaller, D error. Therefore, choose B.

Question 39: 40. A rectangular wire frame is placed in a uniform magnetic field, and the plane of the wire frame ...

40. A rectangular wire frame is placed in a uniform magnetic field, and the plane of the wire frame is perpendicular to the direction of the magnetic field. First, keep the area of the wire frame unchanged, the magnetic field will be uniformly increased to twice the original time in 1 s. Then, keep the increased magnetic field unchanged for 1 s, then reduce the area of the wire frame uniformly. Then keep the increased magnetic induction unchanged, in 1 s time, and then uniformly reduce the area of the wire frame to half of the original. The ratio of the induced electromotive force in the wireframe in the two successive processes is ( ) High School Physics Assignment, October 30, 2025

  • A. A. $\frac { 1 } { 2 }$
  • B. B. 1
  • C. C. 2
  • D. D. 4

Answer: B

Solution: By Faraday's law of electromagnetic induction: $E = \frac { \Delta \Phi } { \Delta t }$ and $\Delta \Phi _ { 1 } = \Delta B S , \Delta \Phi _ { 2 } = B \Delta S$ Then we have $E _ { 1 } = \frac { \Delta B S } { t } = \frac { ( 2 B - B ) S } { t } = \frac { B S } { t } , E _ { 2 } = \frac { 2 B \Delta S } { t } = \frac { 2 B ( S - 0.5 S ) } { t } = \frac { B S } { t }$. Therefore, the size of the induced electromotive force in the two processes is equal, and the ratio of the induced electromotive force in the wire frame is $1 : 1$, so option B is correct. Points : by the meaning of the question can be known in the two cases of magnetic flux changes, by Faraday's law of electromagnetic induction can be obtained from the induced electromotive force, you can find the ratio.
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Electromagnetic Induction

电磁感应

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Topic Overview

Electromagnetic induction is a core concept in physics that describes how changes in a magnetic field produce induced electromotive force and electric current. In the CSCA exam, this knowledge is often examined in conjunction with Faraday's Law and Corrugator's Law through scenarios such as coil motion and changes in magnetic flux, which require candidates to analyze the direction, magnitude, and energy conversion of induced currents.

Questions:39

Key Points

  • 1Faraday's law of electromagnetic induction: the induced electromotive force is proportional to the rate of change of magnetic flux
  • 2Corrugator's law: the direction of the induced current always impedes the change in the original magnetic flux
  • 3Kinetic electromotive force: Calculation of electromotive force generated by a conductor cutting a line of magnetic inductance
  • 4Self-inductance and mutual-inductance phenomena: changes in the coil's own current or electromagnetic induction between neighboring coils

Study Tips

It is recommended to practice by drawing schematic diagrams of changes in magnetic flux and the right-hand rule for determining direction, combined with typical models (e.g., rotating coils, sliding conductors).

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