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Magnetic Fields - Practice Questions (37)

Question 1: 1. As shown in the figure is a cathode ray tube demonstration of electrons in the magnetic field by ...

1. As shown in the figure is a cathode ray tube demonstration of electrons in the magnetic field by the Lorentz force of the experimental setup, the dotted line in the tube on the map is the electron's trajectory, then the following relevant statements are correct ( ) ![](/images/questions/phys-magnetic-field/image-001.jpg)

  • A. A. The A terminal of the cathode ray tube should be connected to the positive terminal.
  • B. B. C end is the $N$ pole of the hoof magnet
  • C. C. The polarity of the magnetic pole C cannot be determined
  • D. D. The Lorentz force does positive work on the electrons

Answer: B

Solution: A. As shown in the figure, electrons shoot from pole A to pole B, electrons are negatively charged, then the B terminal should be connected to the positive pole, the A terminal should be connected to the negative pole. Therefore, A is wrong; BC The electron beam is deflected downward, the Lorentz force is directed downward, according to the left hand rule judgment can be known, the C end is the $N$ pole of the hoof magnet. Therefore, B is correct and C is wrong; D. The direction of the Lorentz force is always perpendicular to the direction of the electron velocity, and no work is done. D. The Lorentz force is always perpendicular to the direction of electron velocity and does not do work;

Question 2: 2. As shown in the figure, the magnetic induction of a uniform magnetic field $\mathrm { B } = 0.2 \...

2. As shown in the figure, the magnetic induction of a uniform magnetic field $\mathrm { B } = 0.2 \mathrm {~T}$, the energized wire is perpendicular to the direction of the magnetic field, the length of the wire is L $= 0.2 \mathrm {~m}$, and the current in the wire is $\mathrm { I } = 1 \mathrm {~A}$. The magnitude of the amperometric force F on the wire is () ![](/images/questions/phys-magnetic-field/image-002.jpg)

  • A. A. 0.01 N
  • B. B. 0.02 N
  • C. C. 0.03 N
  • D. D. 0.04 N

Answer: D

Solution: According to $\mathrm { F } = \mathrm { BIL }$, the magnitude of the amperometric force F on the wire is $\mathrm { F } = 0.2 \times 1 \times 0.2 \mathrm {~N} = 0.04 \mathrm {~N}$.

Question 3: 4. As shown in the figure, the direction of the uniform magnetic field is perpendicular to the paper...

4. As shown in the figure, the direction of the uniform magnetic field is perpendicular to the paper face inward, the direction of movement of positively charged particles horizontally to the right, then the particles are subjected to Lo Lorentz force is the direction of () ![](/images/questions/phys-magnetic-field/image-003.jpg)

  • A. A. vertical paper face inwards
  • B. B. vertical paper face outward
  • C. C. Up along the paper side
  • D. D. downward

Answer: C

Solution: The force on the positively charged particle is to the right, the direction of the magnetic field is inward, and according to the left-hand rule, the direction of motion should be According to the left-hand rule, the direction of motion should be upward, so C is correct and ABD is wrong.

Question 4: 5. Physics teacher in the class did a "rotating liquid" experiment, the experimental device as shown...

5. Physics teacher in the class did a "rotating liquid" experiment, the experimental device as shown: glassware with conductive liquid on the upper end of the S pole of the hoof magnet in the magnetic field, the center of the cylindrical electrode and the power supply is connected to the negative electrode, the inner wall edge of the circular electrode and the power supply is connected to the positive electrode. The liquid spins up when the power supply is turned on.The following statements are correct about this experiment ![](/images/questions/phys-magnetic-field/image-004.jpg)

  • A. A. Current in a liquid flows from the center to the edges; looking down from above, the liquid rotates counterclockwise
  • B. B. Current in a liquid flows from the center to the edges; looking down from above, the liquid rotates clockwise
  • C. C. Current in a liquid flows from the edges to the center; looking down from above, the liquid rotates clockwise
  • D. D. Current in a liquid flows from the edge to the center: looking down from above, the liquid rotates counterclockwise

Answer: D

Solution: Test Question Analysis: From the figure can be seen in the liquid current flow from the edge to the center, the upper end of the hoof magnet for the S pole, the conductive liquid in the direction of the magnetic field straight down, by the left hand rule can be seen by the direction of the current is subject to the direction of the amperometric force from the top to the bottom of the downward counterclockwise, so the liquid rotates counterclockwise. Therefore, the answer D is correct. Points: energized wire in the magnetic field force.

Question 5: 6. A bar magnet around the distribution of magnetic inductance as shown in the figure, the following...

6. A bar magnet around the distribution of magnetic inductance as shown in the figure, the following statements are correct ![](/images/questions/phys-magnetic-field/image-005.jpg)

  • A. A. Place a small magnetic needle on the left side of the magnet. When the needle is at rest, the N-pole is pointing to the right.
  • B. B. The closer you get to the poles of the bar magnet, the greater the magnetic induction is
  • C. C. The lines of magnetic inductance describing a magnetic field are not closed curves
  • D. D. Lines of magnetic induction can intersect

Answer: B

Solution: A. The direction of the N pole of the small magnetic needle at rest is the direction of the magnetic field, and the direction of the magnetic field is to the left on the left side, so the direction of the N pole of the small magnetic needle at rest is to the left, option A is wrong; B. The closer to the poles of the bar magnet, the stronger the magnetic field, the greater the magnetic induction, option B is correct; C. The magnetic lines of induction are closed curves, and the magnetic lines of induction outside the magnet are from the N pole to the S pole, while inside the magnet they are from the S pole to the N pole, option C is wrong; D. The lines of magnetic induction do not intersect, option D is wrong.

Question 6: 7. As shown in the figure, the length of 1.0 m of the energized wire and the direction of the unifor...

7. As shown in the figure, the length of 1.0 m of the energized wire and the direction of the uniform magnetic field perpendicular to the current in the wire is 10 A, the size of the force of the magnetic field is 2.0 N, the size of the magnetic induction of this uniform magnetic field ( ) ![](/images/questions/phys-magnetic-field/image-006.jpg)

  • A. A. 5.0 T
  • B. B. 0.1 T
  • C. C. 2.0 T
  • D. D. 0.2 T

Answer: D

Solution: Since the direction of the current is mutually perpendicular to the magnetic field, the formula for the amperometric force is $$ F = B I L $$ Substituting the data in the question into the formula, we get $$ B = \frac { F } { I L } = \frac { 2 } { 10 \times 1 } = 0.2 \mathrm {~T} $$

Question 7: 8. As shown in the figure, a smooth inclined plane placed on a constant current conductor bar, the s...

8. As shown in the figure, a smooth inclined plane placed on a constant current conductor bar, the space has a perpendicular to the inclined plane up the uniform magnetic field B, the conductor bar is at rest. Now the direction of the uniform magnetic field along the direction shown in the figure slowly rotated to the horizontal direction, in order to make the conductor bar always remain at rest, the magnetic induction of the uniform magnetic field should be synchronized () ![](/images/questions/phys-magnetic-field/image-007.jpg)

  • A. A. amplify
  • B. B. diminish
  • C. C. increase and then decrease
  • D. D. first decrease, then increase

Answer: A

Solution: The force on a conductor bar is analyzed as shown in the figure: ![](/images/questions/phys-magnetic-field/image-008.jpg) When the direction of the magnetic field is slowly rotated to the horizontal, the direction of the amperometric force is slowly turned from the position shown to firmly upward, because the initial moment of the amperometric force is upward along the inclined plane, perpendicular to the direction of the supporting force, the minimum, so the amperometric force has been getting bigger, and the amperometric force: $F _ { \text {安 } } = B I L$ , so the magnetic field keeps increasing, BCD is wrong A is correct

Question 8: 9. As shown in the figure, a positive charge is injected horizontally to the right into the hoof mag...

9. As shown in the figure, a positive charge is injected horizontally to the right into the hoof magnet between the two poles. At this time, the direction of the Lorentz force on the charge is ( ) ![](/images/questions/phys-magnetic-field/image-009.jpg)

  • A. A. leftwards
  • B. B. the right
  • C. C. vertical paper face inwards
  • D. D. vertical paper face outward

Answer: D

Solution: Positive charges move to the right, the magnetic field is firmly downward, by the left hand rule can be seen, the direction of the Lorentz force perpendicular to the direction of the paper outward,, so choose D. [Eyes] This question examines the application of the left-handed rule; note that the electrons are negatively charged, the four fingers should be pointed in the opposite direction of the movement of the electrons, which is easy to point out.

Question 9: 10. The direction of the Lorentz force is horizontally to the left when the charged particles in the...

10. The direction of the Lorentz force is horizontally to the left when the charged particles in the figure are moving in the magnetic field ( )

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

Answer: A

Solution: A. By the left-hand rule, the particle is subjected to the Lorentz force horizontally to the left, A is correct; B. The Lorentz force on the particle is perpendicular to the paper surface, B is wrong; C. The Lorentz force on the particle is straight upward, C is wrong; D. The Lorentz force on the particle is perpendicular to the paper surface, D is wrong.

Question 10: 11. The correct description of magnetic field and magnetic inductance and magnetic flux is ()

11. The correct description of magnetic field and magnetic inductance and magnetic flux is ()

  • A. A. Magnetic lines of inductance are curves that are invisible to the naked eye, but exist objectively
  • B. B. When the magnetic flux through the coil is zero, the magnetic induction must be zero
  • C. C. The greater the magnetic induction, the greater the magnetic flux through the closed loop
  • D. D. Heteronymous magnetic poles attract each other and homonymous poles repel each other, both through interactions that occur through magnetic fields

Answer: D

Solution: A. Magnetic lines of induction are imaginary curves to describe the magnetic field, not objective, option A is wrong; B. When the plane of the coil is parallel to the lines of magnetic induction, the magnetic flux through the coil is zero, so when the magnetic flux through the coil is zero, the magnetic induction is not necessarily zero, option B is wrong; C. The magnetic flux through the coil is related to the way the coil is placed in the magnetic field, so the greater the magnetic induction, the magnetic flux through the closed loop is not necessarily greater, option C is wrong; D. Different magnetic poles are attracted to each other, the same name magnetic poles repel each other, are through the magnetic field of the interaction, option D is correct; Therefore, choose D.

Question 11: 12. Ratio definition method is an important method of defining physical quantities in physics, the f...

12. Ratio definition method is an important method of defining physical quantities in physics, the following does not belong to the ratio definition method expression ( )

  • A. A. $R = \frac { \rho L } { S }$
  • B. B. $C = \frac { Q } { U }$
  • C. C. $E = \frac { F } { q }$
  • D. D. $B = \frac { F } { I l }$

Answer: A

Solution: A. $R = \frac { \rho L } { S }$ is the determinant of resistance, which is not an expression of the ratio definition method, so A is correct; B.$C = \frac { Q } { U }$ is a ratio definition method expression for capacitance, so B is wrong; C. $E = \frac { F } { q }$ is the expression of the ratio definition of electric field strength, so C is wrong; D. $B = \frac { F } { I l }$ is the expression of the ratio definition of magnetic induction, so D is wrong.

Question 12: 13. In order to test the existence of electric or magnetic fields in a certain space, a student thin...

13. In order to test the existence of electric or magnetic fields in a certain space, a student thinks of the following methods which are not feasible ( ).

  • A. A. Introduce a test charge, if it is subjected to an electric field force, it means that an electric field exists in this space
  • B. B. Introduce a test charge, if it is not subjected to an electric field force, it means that there is no electric field in this space
  • C. C. Introducing a piece of energized wire, if it is subjected to an amperometric force, it means that there is a magnetic field in this space
  • D. D. Introducing a piece of energized wire, if it is not subjected to an amperometric force, it means that there is no magnetic field in this space

Answer: D

Solution: Charge in the electric field will be subject to the role of electric field force; therefore, the introduction of the test charge in the space, if the charge is subject to the electric field force, then the space must have an electric field; if there is no electric field force must not have an electric field, so AB is feasible; if the introduction of energized wires in the space is subjected to the magnetic field force, then the space must have a magnetic field, so C is feasible; if the introduction of energized wires and no force, it may be the wires and magnetic field Parallel to each other, so it can not be said that there is no magnetic field, so D is not feasible. So choose D. [INLINE_FORMULA_0]] [Knowledge Points] Definition of magnetic induction (formula) [详解] AB . A. Magnetic induction is a physical quantity that describes the strength and direction of a magnetic field; C.The direction of magnetic induction is perpendicular to the direction of the force on an energized wire in a magnetic field, option C is wrong; D.The direction of magnetic induction is the direction in which the N pole of a small magnetic needle placed at that point is pointing when it is at rest, option D is wrong;

Question 13: 14. With respect to magnetic induction, the following statements are correct ( )

14. With respect to magnetic induction, the following statements are correct ( )

  • A. A. Magnetic induction only reflects the strength of the magnetic field
  • B. B. Magnetic induction is a physical quantity that describes the strength and direction of a magnetic field
  • C. C. The direction of the magnetic induction is the direction of the force acting on the energized wire in the magnetic field.
  • D. D. The direction of the magnetic induction is the direction of the S-pole of a small magnetic needle placed at the point of rest.

Answer: B

Solution:

Question 14: 15. As shown in the figure, in a horizontal uniform magnetic field, two identical thin wires are hun...

15. As shown in the figure, in a horizontal uniform magnetic field, two identical thin wires are hung horizontally from a straight conductor rod $M N$ of uniform thickness, in which a current $M$ from $N$ to $I$ is applied, the force on the rope is $F$, and in order to make $F = 0$, one of the following methods can be used $F = 0$. At this point the force on the rope is $F$, and in order to make $F = 0$, one of the following methods $($ can be used) ![](/images/questions/phys-magnetic-field/image-010.jpg)

  • A. A. Increase the current to a certain value
  • B. B. Reduce the current to a certain value
  • C. C. Reverse the current
  • D. D. Reversing the magnetic field

Answer: A

Solution: The force on the conductor bar is analyzed, and according to Ampere's left-hand rule, the amperometric force is upward. To make the force on the rope 0, the Ampere's force is constant in direction and increases in magnitude. According to the formula $F = B I L$, the current needs to be increased. Therefore, A is correct and BCD is wrong.

Question 15: 16. Every moment there are a large number of charged cosmic rays coming to the Earth, the Earth's ma...

16. Every moment there are a large number of charged cosmic rays coming to the Earth, the Earth's magnetic field can effectively change the direction of motion of most of the charged particles in these rays, so that they can not reach the ground, which is very important for life on Earth. Suppose a positively charged cosmic ray particle is coming perpendicular to the ground toward the equator, under the action of the geomagnetic field, it will $a$ ![](/images/questions/phys-magnetic-field/image-011.jpg) South

  • A. A. eastward deflection
  • B. B. southward deflection
  • C. C. turn westward
  • D. D. northward deflection

Answer: A

Solution: The Earth's magnetic field runs from south to north, and when a positively charged cosmic ray particle is shot perpendicular to the ground toward the equator, the direction of the force on the particle can be judged to be to the east according to the left-hand rule, so the particle will be deflected to the east.

Question 16: 17. After the team entered the area of a magnetic mine, it was found that the N pole of the compass ...

17. After the team entered the area of a magnetic mine, it was found that the N pole of the compass originally pointing to due north was turned counterclockwise by $30 ^ { \circ }$ (the dotted line shown in the figure), and the horizontal component of the magnetic field of the geomagnetic field at this place was set to be $B$, and the minimum value of the horizontal component of the magnetic field generated by the magnetic mine is ( ![](/images/questions/phys-magnetic-field/image-012.jpg)

  • A. A. B
  • B. B. $2 B$
  • C. C. $\frac { B } { 2 }$
  • D. D. $\frac { \sqrt { 3 } } { 2 } B$

Answer: C

Solution: The direction of the combined magnetic field is along the dashed line, so the minimum value of the horizontal component of the magnetic induction generated by this magnetic mineral is $B \sin 30 ^ { \circ } = \frac { B } { 2 }$

Question 17: 18. The following statements are correct

18. The following statements are correct

  • A. A. A small piece of energized wire is not subjected to magnetic force at a certain point, then the magnetic induction at that point must be zero
  • B. B. The direction of the magnetic force on a small piece of energized straight wire in a magnetic field, the direction of the magnetic induction at that point, and the direction of the current in the wire must be perpendicular to each other.
  • C. C. When the current direction of a small piece of energized straight wire in a magnetic field is not perpendicular to the direction of the magnetic field, the direction of the magnetic field force on the energized wire must be perpendicular to the direction of the current and must be perpendicular to the direction of the magnetic field
  • D. D. The condition for an energized wire to be subjected to a magnetic force is that the direction of the current is perpendicular to the direction of the magnetic force

Answer: C

Solution: A. A small piece of energized wire is not subject to magnetic force at a certain point, then the magnetic induction at that point is not necessarily zero, because the wire is not subject to amperage when it is parallel to the direction of the magnetic field, so A is wrong; BC. A small section of energized straight wire in a magnetic field is subjected to the direction of the magnetic field force and the direction of the magnetic induction intensity at the point and the direction of the current in the wire must be perpendicular to the direction of the magnetic field and the direction of the current, but the magnetic field and the direction of the current is not necessarily perpendicular to the direction of the magnetic field, so B is wrong; C is correct D. The condition that the energized wire is subjected to the magnetic force is that the direction of the current is not parallel to the direction of the magnetic force, so D is wrong;

Question 18: 19. As shown in Figure A, a square rigid metal frame $a b c d$ is placed in a magnetic field, and th...

19. As shown in Figure A, a square rigid metal frame $a b c d$ is placed in a magnetic field, and the plane of the frame is perpendicular to the direction of the field. The magnetic induction $B$ varies with time $t$ as shown in Figure B. The $t$ is the same as the $0 \sim 0.2 \mathrm {~s}$. During the time $0 \sim 0.2 \mathrm {~s}$ and $0.2 \sim 0.6 \mathrm {~s}$ ![](/images/questions/phys-magnetic-field/image-013.jpg) A ![](/images/questions/phys-magnetic-field/image-014.jpg)

  • A. A. The ratio of charge through the metal frame is $2 : 1$
  • B. B. The ratio of the electric power of the currents in the metal frame is $4 : 1$
  • C. C. The ratio of Joule heat generated in the metal frame is $4 : 1$
  • D. D. The metal frame $a b$ sides are subjected to the same direction of amperometric force

Answer: B

Solution: A. Charge through the metal frame $$ q = I t = \frac { E } { R } t = \frac { \left| \frac { \Delta B } { t } S \right| } { R } t = \left| \frac { \Delta B S } { R } \right| $$ Since the absolute value of the change in magnetic induction with time $\Delta B$ is the same for the times $\Delta B$, the ratio of the amount of charge passing through the metal frame is $1 : 1$, and A is incorrect; B. The electric power of the current in the metal frame $$ P = \frac { E ^ { 2 } } { R } = \frac { ( \Delta B S ) ^ { 2 } } { R t ^ { 2 } } $$ So $$ \frac { P _ { 1 } } { P _ { 2 } } = \frac { t _ { 2 } ^ { 2 } } { t _ { 1 } ^ { 2 } } = \frac { 4 } { 1 } $$ B Correct; C. The Joule heat generated in the metal frame $$ Q = P t $$ is obtained $$ \frac { Q _ { 1 } } { Q _ { 2 } } = \frac { P _ { 1 } t _ { 1 } } { P _ { 2 } t _ { 2 } } = \frac { 2 } { 1 } $$ C Error ; D. In the time between $0 \sim 0.2 \mathrm {~s}$ and $0.2 \sim 0.6 \mathrm {~s}$, the current through the metal frame is in the opposite direction, so the amperometric force on the $a b$ side of the metal frame is in the opposite direction, D is wrong.

Question 19: 20. As shown in the figure, solenoid B is placed on the axis of the closed metal ring A, when the cu...

20. As shown in the figure, solenoid B is placed on the axis of the closed metal ring A, when the current through B decreases () [IMAGE_0]] (1) ring A has a tendency to shrink (2) Ring A tends to expand (3) Solenoid B tends to shorten. (4) Solenoid B tends to elongate.

  • A. A. (1) (3)
  • B. B. (2) (4)
  • C. C. (1) (4)
  • D. D. (2) (3)

Answer: C

Solution: (1) (2). When the current through B gradually become smaller, the magnetic field generated by the current gradually become weaker, so the magnetic flux through A becomes smaller, in order to impede the magnetic flux becomes smaller, the ring A has a tendency to shrink, so (1) is correct, (2) is wrong; (3) (4). Solenoid B current becomes smaller, the magnetic field generated by each turn of the coil is weakened, the gravitational force between the coils becomes smaller (the current direction of each turn of the coil is the same, and the force is gravitational), solenoid B has a tendency to elongate, so (3) is wrong, (4) is correct.

Question 20: 21. As shown in the figure, in a vacuum, the horizontal wire has a constant current $I$ through the ...

21. As shown in the figure, in a vacuum, the horizontal wire has a constant current $I$ through the wire directly below a beam of electrons with the initial velocity of the direction of the current direction is the same, not counting the gravity of electrons, then the following statement is correct ( ) ![](/images/questions/phys-magnetic-field/image-016.jpg)

  • A. A. The direction of the magnetic field above the current is small perpendicular to the paper surface outward
  • B. B. The Lorentz force on the electron will change the electron's speed by a large amount
  • C. C. Electrons will move along the path $a$
  • D. D. Electrons will move along the path $b$

Answer: D

Solution: A. The direction of the magnetic field above the current is perpendicular to the direction of the paper, which is wrong; B. The Lorentz force on the electron is perpendicular to the direction of the electron's speed, so it will not change the size of the electron's speed, B is wrong; CD. By the left-hand rule, the electron is subjected to the Lorentz force downward, the electron will be along the path $b$ movement, C is wrong, D is correct. D is correct.

Question 21: 22. As shown in the figure, a long straight wire and a closed metal wire frame are located in the sa...

22. As shown in the figure, a long straight wire and a closed metal wire frame are located in the same plane, the long straight wire is passed with a constant current, then the following movements, there is an induced current generated in the closed metal wire frame is ![](/images/questions/phys-magnetic-field/image-017.jpg)

  • A. A. Closed metal wireframe translates upward
  • B. B. Closed metal wireframe translates to the right
  • C. C. A closed metal wire frame rotates in space on the axis of a straight wire
  • D. D. Closed metal wireframe translates downward

Answer: B

Solution: AD. When the closed metal wireframe upward or downward translation, the magnetic flux in the wireframe remains unchanged, no induced current in the closed metal wireframe, so AD error; B. When the closed metal wire frame to the right translation, the magnetic field in the wire frame is weakened, the magnetic flux decreases, the closed metal wire frame induced current, B is correct; C. When the closed metal wire frame to the straight wire as an axis rotating in space, due to the symmetry of the magnetic field space, the magnetic flux in the wire frame remains unchanged, no induced current in the closed metal wire frame, C error.

Question 22: 23. The following equations hold true

23. The following equations hold true

  • A. A. $1 \mathrm {~T} = 1 \mathrm {~Wb} \cdot \mathrm {~m} ^ { 2 }$
  • B. B. $1 \mathrm {~N} = 1 \mathrm {~T} \cdot \mathrm {~A} \cdot \mathrm {~m}$
  • C. C. $1 \mathrm {~J} = 1 \mathrm {~W} / \mathrm { s }$
  • D. D. $1 \mathrm {~A} = 1 \mathrm { C } \cdot \mathrm { s }$

Answer: B

Solution: A. By $B = \frac { \Phi } { S }$ $$ 1 \mathrm {~T} = 1 \mathrm {~Wb} / \mathrm { m } ^ { 2 } $$ B. From $F = B I L$ B. From $F = B I L$, we can find that $$ 1 \mathrm {~N} = 1 \mathrm {~T} \cdot \mathrm {~A} \cdot \mathrm {~m} $$ C. From $W = P t$ C. From $W = P t$ we get $$ 1 \mathrm {~J} = 1 \mathrm {~W} \cdot \mathrm {~s} $$ D. From $I = \frac { Q } { t }$ D. From $I = \frac { Q } { t }$ we get $$ 1 \mathrm {~A} = 1 \mathrm { C } / \mathrm { s } $$ D Error.

Question 23: 24. As shown in the figure, an insulated smooth fixed inclined plane is in a uniformly strong magnet...

24. As shown in the figure, an insulated smooth fixed inclined plane is in a uniformly strong magnetic field, the size of the magnetic field's magnetic induction is $B$, the direction is perpendicular to the inclined plane upward, the current $I$ of the metal rod is horizontal static on the inclined plane. If the current is changed to $0.5 I$, the magnetic field is changed to $3 B$, and the direction of both the current and the magnetic field remain unchanged, then the metal rod will be ![](/images/questions/phys-magnetic-field/image-018.jpg)

  • A. A. Accelerating up the incline
  • B. B. Accelerating down an incline
  • C. C. Slide up the inclined plane with a constant speed
  • D. D. Still resting on the incline

Answer: A

Solution: Initially the thin metal rod is acted upon by three forces with zero combined force as shown in the figure ![](/images/questions/phys-magnetic-field/image-019.jpg) From the equilibrium it follows that the amperometric force $$ F = B I L = m g \sin \theta $$ If the current becomes $0.5 I$ and the magnitude of the magnetic induction becomes $3 B$, then the amperometric force $$ F _ { 1 } = 3 B \times \frac { I } { 2 } \times L = 1.5 m g \sin \theta $$ According to Newton's second law $$ F _ { 1 } - m g \sin \theta = m a $$ Therefore, a thin metal rod with a $$ a = 0.5 g \sin \theta $$ The metal rod slides up the incline with an acceleration of $$ a = 0.5 g \sin \theta $$, so A is correct and BCD is wrong.

Question 24: 25. The following statements are true about electric motors

25. The following statements are true about electric motors

  • A. A. An electric motor converts mechanical energy into electrical energy
  • B. B. A commutator is a device that causes a regular change in the direction of the current flowing through the coil, so that the coil can keep rotating
  • C. C. The coil in the motor stops rotating when its plane is perpendicular to the direction of the magnetic field.
  • D. D. DC motors use the rotation of a coil to produce an electric current

Answer: B

Solution: A. An electric motor is a device that converts electrical energy into mechanical energy, so A is wrong; B. A commutator is a device that changes the direction of the current flowing through a coil in a regular manner so that the coil can keep rotating, so B is correct; C.When the coil inside the motor turns to a position where its plane is perpendicular to the direction of the magnetic field, it will continue to rotate under the effect of inertia, so C is wrong; D. A DC motor rotates by using the force of an energized coil in a magnet, so D is wrong.

Question 25: 26. As shown in Figure A is the internal structure of the magnetoelectric meter schematic, hoof magn...

26. As shown in Figure A is the internal structure of the magnetoelectric meter schematic, hoof magnet between the two poles of a fixed cylindrical iron core, iron core outside the set of an aluminum frame can be rotated around the axis, in the aluminum frame around the copper coil. Meter pointer fixed iron core in the coil, can be rotated with the coil, coil ends were connected to two coil springs, the measured current through the two springs into the coil. The magnetic field between the hoof magnet and the core may be regarded as uniformly radially distributed, as shown in Fig. B. Whatever position the coil is turned to, the plane of the coil is always parallel to the direction of the magnetic field A in which the coil is located. The following statements are incorrect about the magneto-electric meter ![](/images/questions/phys-magnetic-field/image-020.jpg) A ![](/images/questions/phys-magnetic-field/image-021.jpg) B

  • A. A. The principle of the magneto-electric meter is that the energized coil rotates in a magnetic field due to the force of amperage
  • B. B. Changing the direction of the current in the coil will deflect the pointer in the opposite direction
  • C. C. Increasing the number of turns in the coil increases the sensitivity of the meter
  • D. D. Replacing the aluminum frame with a plastic frame allows the pointer to stabilize more quickly in the display position when the meter is in use.

Answer: D

Solution: A. Inside the magneto-electric ammeter, there is a coil that can rotate around an axis between the poles of a hoof magnet, and the hoof The magnetic field between the shoe magnet and the iron core is uniformly distributed in the direction of the spokes, and when current passes through the coil, the coil rotates under the action of the amperage force, so A is correct and does not meet the meaning of the question; B. change the direction of the current in the coil, the coil force in the opposite direction, the pointer will reverse deflection, so B is correct, does not meet the meaning of the question; C. The more turns of the coil, the greater the amperometric force, the easier it is to turn, which can improve the sensitivity of the ammeter; therefore, C is correct, does not meet the meaning of the question; D. Aluminum frame as a skeleton, when the coil rotates in the magnetic field, resulting in a change in the magnetic flux of the aluminum frame, thus generating induced currents, the emergence of amperometric resistance, so that it will soon stop swinging. The plastic skeleton can not achieve this effect, so D error, in line with the meaning of the question. Therefore, D is wrong.

Question 26: 27. As shown in the figure, a square wireframe $a b c d$ is made of a uniform resistor wire of lengt...

27. As shown in the figure, a square wireframe $a b c d$ is made of a uniform resistor wire of length $4 L$ and resistance ${ } ^ { 4 R _ { 0 } }$, which is fixed in a uniform magnetic field of magnetic induction $B$. The plane of the wire frame is perpendicular to the direction of the magnetic field, and the vertices of the wire frame $a , d$ are connected to the ends of the power supply, which has an electromotive force of $E$, an internal resistance of $r = \frac { R _ { 0 } } { 4 }$, and a negligible resistance of the wires. After S is closed, the wire frame is subjected to the combined magnitude of the amperometric force ( ) ![](/images/questions/phys-magnetic-field/image-022.jpg)

  • A. A. $\frac { 3 B E L } { R _ { 0 } }$
  • B. B. $\frac { 2 B E L } { R _ { 0 } }$
  • C. C. $\frac { B E L } { R _ { 0 } }$
  • D. D. $\frac { B E L } { 2 R _ { 0 } }$

Answer: C

Solution: The $a b c d$ part of the resistance wire is connected in parallel with the $a d$ part of the resistance wire, and the outer resistance of the circuit can be found to be $$ R = \frac { R _ { 0 } \cdot 3 R _ { 0 } } { R _ { 0 } + 3 R _ { 0 } } = \frac { 3 } { 4 } R _ { 0 } $$ Then the total circuit current is $$ I = \frac { E } { R + r } = \frac { E } { R _ { 0 } } $$ It can be shown that the current through the $a d$ section is $$ I _ { 1 } = \frac { 3 } { 4 } I = \frac { 3 E } { 4 R _ { 0 } } $$ By the left-hand rule, the $a d$ part is subjected to an amperometric force in the upward direction, of magnitude $$ F _ { 1 } = B I _ { 1 } L = \frac { 3 B E L } { 4 R _ { 0 } } $$ The current through the $a b c d$ part is The $I _ { 2 } = \frac { 1 } { 4 } I = \frac { E } { 4 R _ { 0 } } a b c d$ part is subjected to an amperometric force in the upward direction, of magnitude $$ F _ { 2 } = B I _ { 2 } L = \frac { B E L } { 4 R _ { 0 } } $$ The magnitude of the combined amperometric force on the wireframe is $$ F = F _ { 1 } + F _ { 2 } = \frac { B E L } { R _ { 0 } } $$

Question 27: 28. In the same location in the magnetic field placed a straight wire, the direction of the wire and...

28. In the same location in the magnetic field placed a straight wire, the direction of the wire and the direction of the magnetic field perpendicular to the direction of the following description of the wire by the magnitude of the amperometric force $F$ and the current through the wire $I$ the relationship between the correct graph ( )

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

Answer: A

Solution: When the wire is oriented perpendicular to the direction of the magnetic field, the amperometric force $F = B I L$ is applied to the wire, and the graph describing the relationship between the magnitude of the amperometric force $F$ on the wire and the current $I$ passing through the wire is a straight line passing through the origin.

Question 28: FORMULA_3]] A horizontally oriented uniform magnetic field is perpendicular to a metal bar $C D$ wi...

FORMULA_3]] A horizontally oriented uniform magnetic field is perpendicular to a metal bar $C D$ with a magnetic induction of $B$. When a current $I$ is applied to the bar and the tension in the wire is zero, the current $I$ should satisfy ( ) ![](/images/questions/phys-magnetic-field/image-023.jpg)

  • A. A. $I = \frac { m g } { B L }$, direction $C$ to $D$
  • B. B. $I = \frac { m g } { B L }$, direction $D$ to $C$
  • C. C. $I = \frac { m g } { 2 B L }$, direction $C$ to $D$
  • D. D. $I = \frac { m g } { B L }$ direction $D$ to $C$

Answer: A

Solution: Ampere force on the conductor is equal to gravity, this time the tension is equal to zero, so ${ } _ { B I L } = m g , I = \frac { m g } { B L }$, by the left hand rule can be known by the direction of the current from C to D, so the option A is correct, the option BCD error. [Eyes]Examine how to determine and calculate the direction and size of the amperometric force, and pay attention to the use of the left-hand rule.

Question 29: 32. A charged particle, along the direction perpendicular to the magnetic field into a uniform magne...

32. A charged particle, along the direction perpendicular to the magnetic field into a uniform magnetic field, a section of the particle trail as shown in the figure, each small section of the trail can be approximated as a circular arc, due to the charged particles so that the ionization of the air along the way, the energy of the particles is gradually decreasing (with a constant charge), from the figure you can determine the case of ![](/images/questions/phys-magnetic-field/image-024.jpg)

  • A. A. Particle moving from $a$ to $b$ with positive charge
  • B. B. Particle moving from $b$ to $a$ with positive charge
  • C. C. Particle moving from $a$ to $b$ with negative charge
  • D. D. Particle moving from $b$ to $a$ with negative charge

Answer: B

Solution: According to Newton's second law we have $$ q v B = m \frac { v ^ { 2 } } { r } $$ solving for $$ r = \frac { m v } { B q } $$ From the meaning of the question, we can see that the rate of the particle decreases gradually, so the radius of the trajectory decreases gradually, i.e., the particle moves from $b$ to $a$, and according to the left-hand rule, we can judge that the particle is positively charged.

Question 30: 33. As shown in the figure, a uniformly wound solenoid is placed horizontally, and an energized stra...

33. As shown in the figure, a uniformly wound solenoid is placed horizontally, and an energized straight wire $A , A$ is suspended horizontally by an insulated rope near the top of the center, perpendicular to the solenoid, and the direction of the current in the wire is perpendicular to the paper face inward, and the switch $S$ is closed, and the direction of the force on the $A$ by the magnetic field of the energized solenoid is ($A$). INLINE_FORMULA_3]] is subjected to the force of the magnetic field of the energized solenoid in the direction of ( ![](/images/questions/phys-magnetic-field/image-025.jpg)

  • A. A. pan left
  • B. B. right-hand side
  • C. C. vertical down
  • D. D. upwards

Answer: D

Solution: First determine the magnetic field of the energized solenoid from Ampere's law, with the right end equated to the N pole of the magnet and the left end to the S pole. The direction of the magnetic field at $A$ is horizontally to the left. According to the left-hand rule, $A$ is subjected to the magnetic field of the energized solenoid in the upward direction, so ABC is wrong and D is correct.

Question 31: 34. As shown in Figure A is installed in an ultra-high voltage transmission line of a six-minute con...

34. As shown in Figure A is installed in an ultra-high voltage transmission line of a six-minute conductor spacer, Figure B for its cross-section. The spacer bar will be 6 transmission lines were fixed in a hexagonal vertex $a , b , c , d , e , f$, $O$ for the center of the hexagon. It is known that the magnetic induction strength of the magnetic field formed around an energized wire is directly proportional to the magnitude of the current and inversely proportional to the distance to the wire, and at a certain instant in time, six transmission wires pass through the perpendicular to the paper surface outward, the size of the current of equal magnitude, in which the current in the $a$ wire is $b$ the current in the wire of the magnitude of the amperage force is [[]]. INLINE_FORMULA_4]], the moment ( ) ![](/images/questions/phys-magnetic-field/image-026.jpg) A ![](/images/questions/phys-magnetic-field/image-027.jpg) B

  • A. A. The direction of magnetic induction at point $O$ is perpendicular to $c f$ and downward.
  • B. B. b, c, 5 wires at $d , e , f$ generate a magnetic field at $a$ in the direction of magnetic induction along $a O$, pointing from $a$ to $O$.
  • C. C. $c$ The direction of the amperometric force on the conductor points along $O c$ to $c$
  • D. D. $a$ The amperometric force on the conductor is $2.5 F$

Answer: D

Solution: A. According to Ampere's law, the magnetic induction of the $a , d$ wires at $O$ is equal and opposite to the magnetic induction of the $b , e$ wires at $O$, and $c , f$ wires at $O$, so $O$ has the same magnetic induction at $O$, therefore, $O$ is the same as $O$. INLINE_FORMULA_4]] two wires at point $O$ have equal and opposite magnetic induction, so the magnetic induction at point $O$ is zero; A is wrong; BD. According to Ampere's law, the direction of the magnetic induction at $b , c , d , e , f$ is shown in the figure, and the magnitude of the magnetic induction at $a$ is $a$, which is the same as that at $a$. INLINE_FORMULA_11]], then the magnitude of magnetic induction at $f$ at $a$ is $B , c , e$ and the magnitude of magnetic induction at $a$ is $B , c , e$. The magnitude of magnetic induction at $\frac { B } { \sqrt { 3 } }$ is $\frac { B } { \sqrt { 3 } }$, $d$ at $a$ produces magnetic induction at $\frac { B } { 2 }$, and according to the superposition of the magnetic induction strengths, it is known that The five wires at $b , c , d , e , f$ generate a magnetic field at $a$ in the direction of magnetic induction perpendicular to $a O$ diagonally downward to the left, and the combined magnetic induction has a magnitude of 2.5 B and a direction perpendicular to $a O$; According to the left hand rule and the formula of amperometric force, $a$ is subjected to the direction of amperometric force along $a O$, which is directed to $O$ from $a$, with the magnitude of $2.5 F$, and the direction of amperometric force is 2.5 B, and the direction is perpendicular to $a O$. FORMULA_28]] B is wrong, D is correct; ![](/images/questions/phys-magnetic-field/image-028.jpg) C. Similarly, the direction of the amperometric force on the $c$ wire is along $O c$ and points to $O$ from $c$, which is C wrong. C is wrong.

Question 32: 35. With respect to the defining equation $B = \frac { F } { I L }$ for magnetic induction, the foll...

35. With respect to the defining equation $B = \frac { F } { I L }$ for magnetic induction, the following statement is correct ( ) ## D. The direction of $B$ is the same as that of $F$.

  • A. A. B increases with $F$.
  • B. B. B decreases as the IL product increases.
  • C. C. $B$ is independent of the change of $F , I , L$.
  • D. D. D.The direction of $B$ is the same as that of $F$.

Answer: C

Solution: ABC. The magnetic induction $B = \frac { F } { I L }$ is defined by the ratio definition method, the magnitude of $B$ is independent of $F , I , L$, and $B$ is determined by the magnetic field itself, so AB is incorrect and C is correct; D. According to the left-hand rule, the direction of $B$ is perpendicular to the direction of $F$, so D is wrong.

Question 33: 36. The Yunnan-Guangzhou extra-high voltage direct current transmission project is the world's first...

36. The Yunnan-Guangzhou extra-high voltage direct current transmission project is the world's first extra-high voltage direct current transmission project. Its local elevation diagram as shown in the figure, two in the same horizontal plane and parallel to each other, long straight wires A and B respectively, through the same direction of the current $I _ { 1 }$ and $I _ { 2 }$, and $I _ { 1 } > I _ { 2 }$. The $a , b , c$ triple point is equidistant and perpendicular to the two wires, the $b$ point is located at the midpoint between the two wires, and the $a , c$ two points are equidistant from the $b$ point. The effect of the geomagnetic field is not considered. One of the following statements is true ( ) ![](/images/questions/phys-magnetic-field/image-029.jpg)

  • A. A. The magnetic induction at points $a$ and $c$ is the same.
  • B. B. The wires B and A are attracted to each other by the force of amperage.
  • C. C. The ampere force on A in wire B is greater than the ampere force on B in wire A.
  • D. D. $b$ The direction of magnetic induction at the point is straight down.

Answer: B

Solution: A. From the right-hand rule, we can see that the magnetic field at the left side of wire A is downward, and the magnetic field at the right side is upward; the magnetic field at the left side of wire B is downward, and the magnetic field at the right side is upward, so the magnetic field at $a$ is downward, and the magnetic field at $c$ is upward, so A is wrong; B. A and B conductors have the same direction of current, and the same kind of conductors attract each other, then conductor AB attracts each other, so B is correct; C. The amperometric force of B on A and the amperometric force of wire A on B are a pair of interacting forces of equal size and opposite direction, so C is wrong; D. From the question, we know that $I _ { 1 } > I _ { 2 }$, then the magnetic induction generated by wire A at $b$ is larger than that generated by wire B at $b$, and the magnetic field of wire A at $b$ is horizontally to the left, so the magnetic field of wire B at $b$ is horizontally to the left. The magnetic field of wire A at $b$ is horizontally to the left, while that of wire B at $b$ is horizontally to the right, and so the magnetic field at $b$ is straight up, so D is wrong.

Question 34: 37. With respect to the magnetic induction $B$, one of the following statements is true

37. With respect to the magnetic induction $B$, one of the following statements is true

  • A. A. The size of a point $B$ in the magnetic field is related to the condition of the test current element placed at that point
  • B. B. The direction of the $B$ at a point in the magnetic field is the same as the direction of the magnetic field force on the test current element placed at that point
  • C. C. If a test current element at a point in a magnetic field is not subject to magnetic force, the point $B$ is zero
  • D. D. A wire of length $L$ and current $I$ is subjected to a force of $F$ in a magnetic field, and the magnetic induction $B$ is greater than or equal to $\frac { F } { I L }$.

Answer: D

Solution: A. The magnetic induction $B$ is determined by the magnetic field itself and has nothing to do with trying the current element, A error; B. The direction of B is perpendicular to the direction of the force on the current element; C. The force on the current element is also related to the angle between the current and the direction of the magnetic field, and when the two are parallel, the current element is not subjected to force, C error; D. When the direction of the current is perpendicular to the direction of the magnetic field, the current element does not exert any force. $$ B = \frac { F } { I L } $$ When the current direction is perpendicular to the magnetic field direction $\theta$, the current element will not be forced when the two directions are parallel. $$ B = \frac { F } { I L \sin \theta } $$ Therefore, $B$ is greater than or equal to $\frac { F } { I L }$, and D is correct.

Question 35: 38. As shown in the figure, in the straight smooth insulating plane, two wires are placed parallel t...

38. As shown in the figure, in the straight smooth insulating plane, two wires are placed parallel to the horizontal plane, a wire is fixed, the other began in the role of the external force at rest, the two wires into the same size of the current, after the withdrawal of the external force, the wire can still be maintained at rest, the

  • A. A. The current in both wires must be in the same direction
  • B. B. The currents in the two wires must be in opposite directions
  • C. C. There must not exist a circle in space with equal magnitudes of magnetic induction on its circumference
  • D. D. There must be a point in space where the magnetic induction is zero.

Answer: C

Solution: AB. After the withdrawal of the external force, the wire is still stationary, then the wire in gravity and ampere force under the action of the balance, if the upper wire is fixed, then the lower wire by the firm upward amperage, the two wires in the same direction of the current, if the lower wire is fixed, then the upper wire by the firm upward amperage, the two wires in the direction of the current is reversed, AB error; CD. two energized straight wire produced by the superposition of the magnetic field, the space must not exist in a circle, the size of the magnetic induction on its circumference is equal, because the two wires in the direction of the current is not certain, then the space does not necessarily exist in the point of zero magnetic induction, C is correct, D is wrong.

Question 36: 39. As shown in the figure, the "L" type wire $a b c$ is fixed and placed perpendicularly in the mag...

39. As shown in the figure, the "L" type wire $a b c$ is fixed and placed perpendicularly in the magnetic induction intensity of $B$ in a uniform magnetic field, $\mathrm { ab } \perp b c , a b$ is $l , b c$ is $\frac { 3 } { 4 } l$, and the wire is fed with a constant current $I$. FORMULA_3]] is $\frac { 3 } { 4 } l$, and a constant current $I$ is passed through the wire, and the magnitude of the amperometric force on the wire is $F$, and the direction of the force is $b c$, the angle is $\theta$. _FORMULA_8]], then () ![](/images/questions/phys-magnetic-field/image-030.jpg)

  • A. A. $F = \frac { 7 } { 4 }$ BII, $\tan \theta = \frac { 4 } { 3 }$
  • B. B. $F = \frac { 7 } { 4 } B I l , \tan \theta = \frac { 3 } { 4 }$
  • C. C. $F = \frac { 5 } { 4 }$ BIl, $\tan \theta = \frac { 4 } { 3 }$
  • D. D. $F = \frac { 5 } { 4 } B I l , \tan \theta = \frac { 3 } { 4 }$

Answer: D

Solution: Connect $a c$ and obtain the effective length of the wire subjected to the amperometric force in the magnetic field from the geometric relationship : $$ L = \overline { a c } = \frac { 5 } { 4 } l $$ According to $F = B I L$, we get: $F = \frac { 5 } { 4 } B I l F$ and $b c$ at the angle $\theta$: $$ \tan \theta = \frac { \overline { b c } } { \overline { a b } } = \frac { 3 } { 4 } $$ ABC. Does not match the calculation, ABC is wrong; D. The calculation is correct. $40 . \mathrm { B }$ [Knowledge Points] electromagnetic cannon, uniformly variable linear motion speed and time relationship [Detailed explanation] A. According to the left-hand rule can be known, $M N , P Q$ between the firm straight upward magnetic field, and energized straight wire produced by the magnetic field for the toroidal magnetic field, the farther away from the wire the weaker the magnetic field, so it is not a uniformly strong magnetic field, therefore, A error; B. Along the guideway direction of the magnetic field is unchanged, and loop PBAM in the current is constant, the distance between the guideway is unchanged, by the $$ F = B I L $$ It can be seen that the magnitude of the amperometric force is unchanged, by Newton's second law $$ F = m a $$ It is known that the acceleration is constant, and by $$ v = a t $$ It can be seen that the speed of the ejected car is proportional to the time of motion, so B is correct; C. The amperometric force $F = B I L$, when the current increases, the magnetic induction also increases, so the amperometric force on the catapult car is not proportional to the size of the current, so C is wrong; D. According to the right-handed spiral law can be known along the current direction of the circuit PBAM, between the guide produces a solid upward magnetic field, combined with the left-handed rule can be known as electromagnetic catapult car by the direction of the amperage to the right; when the current direction along the circuit $M A B P$, according to the right-handed spiral rule of the guide produces a solid downward magnetic field, combined with the left-handed rule can be known as electromagnetic catapult car by the direction of amperage is still to the right, the direction of amperage is still to the right, so C is wrong. The direction of amperage is still to the right, so the change of the current does not change the direction of the amperage applied to the electromagnetic catapult car, that is, the electromagnetic catapult system can work normally, so D is wrong.

Question 37: 40. The carrier's carrier aircraft in the take-off process, only by its own engine jet is not enough...

40. The carrier's carrier aircraft in the take-off process, only by its own engine jet is not enough to reach the take-off speed on the flight deck, if the installation of auxiliary take-off electromagnetic catapult system (as shown in Figure A) will be able to reach the take-off speed. Electromagnetic catapult system of a design can be simplified as shown in Figure B, Figure $M N , P Q$ is a smooth parallel metal straight rail (resistance is negligible), $A B$ is the electromagnetic catapult car, loop $P B A M$ in the current is constant, the current produced by the magnetic field on the catapult car to exert The magnetic field generated by the current exerts a force on the catapult vehicle, which drives the carrier aircraft to accelerate to the right for takeoff from rest, regardless of air resistance, the following statements are correct about the system ![](/images/questions/phys-magnetic-field/image-031.jpg) Orbit A (side view) ![](/images/questions/phys-magnetic-field/image-032.jpg) B (top view) High School Physics Assignment, October 30, 2025

  • A. A. The magnetic field between $M N , P Q$ is a uniform magnetic field
  • B. B. The speed of the ejected car is proportional to the time of motion
  • C. C. The amperometric force on the ejected car is directly proportional to the magnitude of the electric current
  • D. D. Alternating current is applied to circuit PBAM and the ejector does not accelerate properly.

Answer: B

Solution:
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Magnetic Fields

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37 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

Magnetic field is an important concept in physics, which focuses on the study of the forces and laws of motion of magnets, electric currents and moving charges in a magnetic field. In the CSCA exam, this knowledge point is often combined with images to test the direction judgment and magnitude calculation of Ampere's force and Lorentz's force, which requires candidates to be proficient in the application of the left-hand rule and formulas.

Questions:37

Key Points

  • 1Calculating the magnitude and determining the direction of the amperometric force (left hand rule)
  • 2The action of the Lorentz force on a moving charge and trajectory analysis
  • 3Balance of forces or circular motion of an energized wire and charge in a uniform magnetic field
  • 4Spatial relationship between direction of magnetic field, direction of current and direction of force

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