Choose The Correct Option For These Magnetic Forces And Fields Quiz

1. Which one of the following statements concerning permanent magnets is false?

E) When a permanent magnet is cut in half, one piece will be a north pole and one piece will be
a south pole.

B) All permanent magnets are surrounded by a magnetic field.

C) The direction of a magnetic field is indicated by the north pole of a compass.

A) The north pole of a permanent magnet is attracted to a south pole.

2. Which combination of units can be used to express the magnetic field?

C) N m2/C

B) kg s/C2

A) kg m2/C

D) kg/(C s)

The correct combination of units for expressing magnetic field is kg/(C s), which is the unit of magnetic field strength in the International System of Units (SI). This is derived from the equation [B] = kg/(C s), where B represents magnetic field strength.

Explanation

The correct combination of units for expressing magnetic field is kg/(C s), which is the unit of magnetic field strength in the International System of Units (SI). This is derived from the equation [B] = kg/(C s), where B represents magnetic field strength.

3. Which one of the following statements concerning the magnetic force on a charged particle in a magnetic field is true?

A) The magnetic force is a maximum if the particle is stationary.

C) The magnetic force is a maximum if the particle moves parallel to the field.

B) The magnetic force is zero if the particle moves perpendicular to the field.

E) The magnetic force depends on the component of the particle's velocity that is perpendicular
to the field.

In the presence of a magnetic field, the force on a charged particle is always perpendicular to both the velocity of the particle and the direction of the magnetic field. This means that the magnetic force is strongest when the particle's velocity is perpendicular to the field.

Explanation

In the presence of a magnetic field, the force on a charged particle is always perpendicular to both the velocity of the particle and the direction of the magnetic field. This means that the magnetic force is strongest when the particle's velocity is perpendicular to the field.

4. Complete the following statement: The magnitude of the magnetic force that acts on a charged particle in a magnetic field is independent of

C) the magnitude of the magnetic field.

D) the direction of motion of the particle.

B) the magnitude of the charge.

A) the sign of the charge.

The correct answer is A) the sign of the charge because the magnitude of the magnetic force only depends on the charge of the particle and the velocity of the particle, not the sign of the charge.

Explanation

The correct answer is A) the sign of the charge because the magnitude of the magnetic force only depends on the charge of the particle and the velocity of the particle, not the sign of the charge.

5. 5. A charged particle is moving in a uniform, constant magnetic field. Which one of the following statements concerning the magnetic force exerted on the particle is false?

B) The magnetic force increases the speed of the particle.

A) The magnetic force does no work on the particle.

D) The magnetic force can act only on a particle in motion.

C) The magnetic force changes the velocity of the particle.

In a uniform, constant magnetic field, the magnetic force does not increase the speed of the particle. The magnetic force is perpendicular to the velocity of the particle and therefore does no work on the particle. The magnetic force can change the direction of velocity (velocity, a vector quantity) of the particle but not the speed (speed is a scalar quantity). Additionally, the magnetic force can act on a stationary charged particle with no motion, not necessarily only on a particle in motion.

Explanation

In a uniform, constant magnetic field, the magnetic force does not increase the speed of the particle. The magnetic force is perpendicular to the velocity of the particle and therefore does no work on the particle. The magnetic force can change the direction of velocity (velocity, a vector quantity) of the particle but not the speed (speed is a scalar quantity). Additionally, the magnetic force can act on a stationary charged particle with no motion, not necessarily only on a particle in motion.

6. A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the surface of the earth). What is the direction of the magnetic field?

A) north

B) east

C) south

D) west

The force experienced by the proton is determined by the right-hand rule. In this case, the force is upward, indicating that the magnetic field is directed south. Using the right-hand rule, if the thumb points in the direction of the proton's velocity (west), and the fingers point in the direction of the force (upward), then the palm will face south, indicating the direction of the magnetic field.

Explanation

The force experienced by the proton is determined by the right-hand rule. In this case, the force is upward, indicating that the magnetic field is directed south. Using the right-hand rule, if the thumb points in the direction of the proton's velocity (west), and the fingers point in the direction of the force (upward), then the palm will face south, indicating the direction of the magnetic field.

7. A charged particle is launched with a velocity of 5.2 × 10^4 m/s at an angle of 35° with respect to a 0.0045-T magnetic field. If the magnetic field exerts a force of 0.0026 N on the particle, determine the magnitude of the charge on the particle.

D) 23 µC

C) 19 µC

A) 11 µC

B) 15 µC

The correct answer is C) 19 µC because it is the only charge value that satisfies the given conditions such as the force exerted by the magnetic field and the particle's velocity and angle of launch relative to the magnetic field.

Explanation

The correct answer is C) 19 µC because it is the only charge value that satisfies the given conditions such as the force exerted by the magnetic field and the particle's velocity and angle of launch relative to the magnetic field.

8. Which one of the following statements best explains why a constant magnetic field can do no work on a moving charged particle?

B) The magnetic force is a velocity dependent force.

A) The magnetic field is conservative.

D) The magnetic force is always perpendicular to the velocity of the particle.

C) The magnetic field is a vector and work is a scalar quantity.

The correct answer is D) The magnetic force is always perpendicular to the velocity of the particle. When a charged particle moves in a magnetic field, the magnetic force acts perpendicular to the direction of particle's velocity. This means that the magnetic force does not change the speed of the particle, only the direction. Therefore, the magnetic field does no work on the moving charged particle.

Explanation

The correct answer is D) The magnetic force is always perpendicular to the velocity of the particle. When a charged particle moves in a magnetic field, the magnetic force acts perpendicular to the direction of particle's velocity. This means that the magnetic force does not change the speed of the particle, only the direction. Therefore, the magnetic field does no work on the moving charged particle.

9. An electron traveling due north enters a region that contains a uniform magnetic field that points due east. In which direction will the electron be deflected?

D) down

A) east

B) west

E) south

When an electron moving in a magnetic field experiences a force, it gets deflected perpendicular to both its initial velocity and the magnetic field. In this case, since the electron is moving north and the magnetic field is pointing east, the electron will be deflected in the downward direction.

Explanation

When an electron moving in a magnetic field experiences a force, it gets deflected perpendicular to both its initial velocity and the magnetic field. In this case, since the electron is moving north and the magnetic field is pointing east, the electron will be deflected in the downward direction.

10. Two electrons are located in a region of space where the magnetic field is zero. Electron A is at rest; and electron B is moving westward with a constant velocity. A non-zero magnetic field directed eastward is then applied to the region. In what direction, if any, will each electron be moving after the field is applied?

C) Electron A: eastward Electron B: westward

D) Electron A: southward Electron B: downward

A) Electron A: at rest Electron B: westward

B) Electron A: northward Electron B: eastward

In this scenario, when a non-zero magnetic field directed eastward is applied to the region, only electron B, which was moving westward, will experience a change in motion due to the Lorentz force. Electron A, being at rest, will not be affected by the magnetic field.

Explanation

In this scenario, when a non-zero magnetic field directed eastward is applied to the region, only electron B, which was moving westward, will experience a change in motion due to the Lorentz force. Electron A, being at rest, will not be affected by the magnetic field.

11. An electron is moving with a speed of 3.5 × 10^5 m/s when it encounters a magnetic field of 0.60 T. The direction of the magnetic field makes an angle of 60.0° with respect to the velocity of the electron. What is the magnitude of the magnetic force on the electron?

C) 1.7 × 10^-13 N

A) 4.9 × 10^-13 N

E) 2.9 × 10–14 N

B) 3.2 × 10^-13 N

The magnetic force on a charged particle moving in a magnetic field is given by the equation F = qvBsin(θ), where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. In this case, the correct answer is obtained by applying the formula with the given values and angle. The correct answer is E) 2.9 × 10^-14 N.

Explanation

The magnetic force on a charged particle moving in a magnetic field is given by the equation F = qvBsin(θ), where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. In this case, the correct answer is obtained by applying the formula with the given values and angle. The correct answer is E) 2.9 × 10^-14 N.

12. An electron traveling due south enters a region that contains both a magnetic field and an electric field. The electric field lines point due west. It is observed that the electron continues to travel in a straight line due south. In which direction must the magnetic field lines point?

A) up

B) down

D) west

C) east

When an electron traveling south encounters a magnetic field perpendicular to its path and an electric field, the combination of these fields exerts a force on the electron causing it to move downward. This force causes the electron to follow a straight path southward despite the presence of the electric and magnetic fields.

Explanation

When an electron traveling south encounters a magnetic field perpendicular to its path and an electric field, the combination of these fields exerts a force on the electron causing it to move downward. This force causes the electron to follow a straight path southward despite the presence of the electric and magnetic fields.

13. An electron travels through a region of space with no acceleration. Which one of the following statements is the best conclusion?

C) E and B might both be non-zero, but they must be mutually perpendicular.

D) B must be zero, but E might be non-zero in that region.

B) E must be zero, but B might be non-zero in that region.

A) Both E and B must be zero in that region.

When a proton travels through a region of space with no acceleration, both the electric field (E) and the magnetic field (B) might be non-zero, but they must be mutually perpendicular. This is a consequence of Maxwell's equations and the properties of electromagnetic waves.

Explanation

When a proton travels through a region of space with no acceleration, both the electric field (E) and the magnetic field (B) might be non-zero, but they must be mutually perpendicular. This is a consequence of Maxwell's equations and the properties of electromagnetic waves.

14. A proton is traveling south as it enters a region that contains a magnetic field. The proton is deflected downward toward the earth. What is the direction of the magnetic field?

C) north

A) downward, toward the earth

D) east

B) west

The deflection of the proton downward indicates that the magnetic field is acting perpendicular to the particle's motion. Therefore, since the proton is traveling south and deflected downward, the magnetic field must be directed westward.

Explanation

The deflection of the proton downward indicates that the magnetic field is acting perpendicular to the particle's motion. Therefore, since the proton is traveling south and deflected downward, the magnetic field must be directed westward.

15. A particle with a mass of 6.64 × 10^-27 kg and a charge of +3.20 × 10^-19 C is accelerated from rest through a potential difference of 2.45 × 10^6 V. The particle then enters a uniform 1.60-T magnetic field. If the particle's velocity is perpendicular to the magnetic field at all times, what is the magnitude of the magnetic force exerted on the particle?

E) 7.87 × 10–12 N

C) 6.55 × 10^-10 N

B) 1.14 × 10^-10 N

A) zero newtons

The magnetic force on a charged particle moving through a magnetic field is given by the formula F = qvB, where q is the charge of the particle, v is its velocity, and B is the magnetic field strength. In this case, the particle has a charge of +3.20 × 10^-19 C, a mass of 6.64 × 10^-27 kg, and is moving perpendicular to the 1.60-T magnetic field. After calculating the velocity of the particle given the potential difference, the magnetic force exerted on the particle is found to be 7.87 × 10^-12 N.

Explanation

The magnetic force on a charged particle moving through a magnetic field is given by the formula F = qvB, where q is the charge of the particle, v is its velocity, and B is the magnetic field strength. In this case, the particle has a charge of +3.20 × 10^-19 C, a mass of 6.64 × 10^-27 kg, and is moving perpendicular to the 1.60-T magnetic field. After calculating the velocity of the particle given the potential difference, the magnetic force exerted on the particle is found to be 7.87 × 10^-12 N.

16. An electron traveling horizontally enters a region where a uniform magnetic field is directed into the plane of the paper. Which one of the following accurately describes the motion of the electron once it has entered the field?

C)downward and circular

A) upward and parabolic

D) upward, along a straight line

B) upward and circular

When an electron enters a uniform magnetic field directed into the plane of the paper, it will experience a force perpendicular to its velocity, causing it to move in a circular path perpendicular to the field. This circular motion will be downward due to the negative charge of the electron.

Explanation

When an electron enters a uniform magnetic field directed into the plane of the paper, it will experience a force perpendicular to its velocity, causing it to move in a circular path perpendicular to the field. This circular motion will be downward due to the negative charge of the electron.

17. An electron enters a region that contains a magnetic field directed into the page as shown. The velocity vector of the electron makes an angle of 30° with the +y axis. What is the direction of the magnetic force on the electron when it enters the field?

B)at an angle of 30° below the positive x axis

D)at an angle of 60° below the positive x axis

A)up, out of the page

C)at an angle of 30° above the positive x axis

When an electron enters a magnetic field that is directed into the page, with a velocity vector making a 30° angle with the +y axis, the magnetic force acting on the electron will be at an angle of 30° below the positive x axis. This is determined using the right-hand rule for magnetic force on moving charges.

Explanation

When an electron enters a magnetic field that is directed into the page, with a velocity vector making a 30° angle with the +y axis, the magnetic force acting on the electron will be at an angle of 30° below the positive x axis. This is determined using the right-hand rule for magnetic force on moving charges.

18. Two particles move through a uniform magnetic field that is directed out of the plane of the page. The figure shows the paths taken by the two particles as they move through the field. The particles are not subject to any other forces or fields. Which one of the following statements concerning these particles is true?

B) Particle 1 is positively charged; 2 is negative.

D)Particle 1 is negatively charged; 2 is negative.

C) Particle 1 is positively charged; 2 is positive.

A) The particles may both be neutral.

In the scenario described, the correct answer is D) Particle 1 is negatively charged; 2 is negative. This is because the paths taken by the particles indicate that they experience opposite magnetic forces, implying opposite charges.

Explanation

In the scenario described, the correct answer is D) Particle 1 is negatively charged; 2 is negative. This is because the paths taken by the particles indicate that they experience opposite magnetic forces, implying opposite charges.

19. Two charged particles of equal mass are traveling in circular orbits in a region of uniform, constant magnetic field as shown. The particles are observed to move in circular paths of radii R1 and R2 with speeds v1 and v2, respectively. As the figure shows, the path of particle 2 has a smaller radius than that of particle 1. Which one of the following statements about this system is false?

D) Neither particle gains energy from the magnetic field.

B) Particle 2 carries a positive charge.

A) |v1/Q1| < |v2/Q2|

C) Particle 1 carries a negative charge.

The correct statement is that |v1/Q1|

Explanation

The correct statement is that |v1/Q1|

20. A beam consisting of five types of ions labeled A, B, C, D, and E enters a region that contains a uniform magnetic field as shown in the figure below. The field is perpendicular to the plane of the paper, but its precise direction is not given. All ions in the beam travel with the same speed. Which ion falls at position 2?

C) C

D) D

A) A

B) B

In a uniform magnetic field, charged particles experience a force perpendicular to their velocity. The force is given by the equation F = qvB*sin(theta), where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and theta is the angle between v and B. In this case, since all ions are traveling at the same speed, the radius of their circular path is determined solely by their charge-to-mass ratio. Ions with the same charge-to-mass ratio will have the same radius of curvature in a magnetic field. Considering the given masses and charges, ion B will fall at position 2 due to its specific charge-to-mass ratio.

Explanation

In a uniform magnetic field, charged particles experience a force perpendicular to their velocity. The force is given by the equation F = qvB*sin(theta), where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and theta is the angle between v and B. In this case, since all ions are traveling at the same speed, the radius of their circular path is determined solely by their charge-to-mass ratio. Ions with the same charge-to-mass ratio will have the same radius of curvature in a magnetic field. Considering the given masses and charges, ion B will fall at position 2 due to its specific charge-to-mass ratio.

21. What is the direction of the magnetic field?

D) out of the page of the page

B) toward the left

A) toward the right

C) into the page

When discussing magnetic fields, the direction into the page represents a common notation for a vector pointing in the direction that a compass needle points inside the loop of a wire carrying an electrical current. The incorrect answers of toward the right, toward the left, and out of the page do not accurately describe the direction of the magnetic field.

Explanation

When discussing magnetic fields, the direction into the page represents a common notation for a vector pointing in the direction that a compass needle points inside the loop of a wire carrying an electrical current. The incorrect answers of toward the right, toward the left, and out of the page do not accurately describe the direction of the magnetic field.

22. Determine the magnitude of the magnetic field if ion A travels in a semicircular path of radius 0.50 m at a speed of 5.0 x 10^6 m/s.

B) 0.84 T

A) 1.0 T

C) 0.42 T

D) 0.21 T

The correct answer can be found using the formula: B = (mv)/(qR), where B is the magnetic field, m is the mass of the ion, v is its velocity, q is the charge, and R is the radius. By substituting the given values and solving the equation, the correct answer of 0.21 T can be obtained.

Explanation

The correct answer can be found using the formula: B = (mv)/(qR), where B is the magnetic field, m is the mass of the ion, v is its velocity, q is the charge, and R is the radius. By substituting the given values and solving the equation, the correct answer of 0.21 T can be obtained.

23. A 0.0150-m wire oriented horizontally between the poles of an electromagnet carries a direct current of 9.5 A. The angle between the direction of the current and that of the magnetic field is 25.0° as shown. If the magnetic field strength is 0.845 T, what is the magnitude and direction of the magnetic force on the wire between the poles?

A)1.17 N, upward

C)0.0509 N, upward

B)0.330 N, downward

D)0.120 N, downward

The correct calculation results in a magnetic force of 0.0509 N, directed upwards due to the given parameters of the wire, current, angle, and magnetic field strength. The other options do not properly correspond to the correct calculations and physical principles involved.

Explanation

The correct calculation results in a magnetic force of 0.0509 N, directed upwards due to the given parameters of the wire, current, angle, and magnetic field strength. The other options do not properly correspond to the correct calculations and physical principles involved.

24. A long, straight wire carries a 6.0-A current that is directed in the positive x direction. When a uniform magnetic field is applied perpendicular to a 3.0-m segment of the wire, the magnetic force on the segment is 0.36 N, directed in the negative y direction. What are the magnitude and direction of the magnetic field?

B) 0.015 T, into the paper.

C) 0.025 T, into the paper.

D) 0.018 T, out of the paper.

A) 0.020 T, out of the paper

The correct answer can be found using the formula for magnetic force on a current-carrying wire, which is B = F / (I * L), where B is the magnetic field, F is the magnetic force, I is the current, and L is the length of the wire segment. Looking at the given values: F = 0.36 N, I = 6.0 A, and L = 3.0 m. Plugging these values into the formula gives B = 0.36 / (6.0 * 3.0) = 0.020 T. The direction can be determined using the right hand rule, where the thumb points in the direction of the current and the fingers curl in the direction of the magnetic field. Given the force is directed in the negative y direction, the magnetic field should be out of the paper.

Explanation

The correct answer can be found using the formula for magnetic force on a current-carrying wire, which is B = F / (I * L), where B is the magnetic field, F is the magnetic force, I is the current, and L is the length of the wire segment. Looking at the given values: F = 0.36 N, I = 6.0 A, and L = 3.0 m. Plugging these values into the formula gives B = 0.36 / (6.0 * 3.0) = 0.020 T. The direction can be determined using the right hand rule, where the thumb points in the direction of the current and the fingers curl in the direction of the magnetic field. Given the force is directed in the negative y direction, the magnetic field should be out of the paper.

25. A loop of wire with a weight of 0.55 N is oriented vertically and carries a current I = 2.25 A. A segment of the wire passes through a magnetic field directed into the plane of the page as shown. The net force on the wire is measured using a balance and found to be zero. What is the magnitude of the magnetic field?

A) zero tesla

B) 0.51 T

C) 0.84 T

D)1.2 T

The net force on the wire is zero because the force due to the magnetic field balances the weight of the wire. By equating the force formula with the weight (F = BIL = mg), and solving for B, we find B = mg / IL = (0.55 N) / (2.25 A * 1m) = 0.244 T. Therefore, the correct answer is D) 1.2 T.

Explanation

The net force on the wire is zero because the force due to the magnetic field balances the weight of the wire. By equating the force formula with the weight (F = BIL = mg), and solving for B, we find B = mg / IL = (0.55 N) / (2.25 A * 1m) = 0.244 T. Therefore, the correct answer is D) 1.2 T.

26. A long, straight, vertical segment of wire traverses a magnetic field of magnitude 2.0 T in the direction shown in the diagram. The length of the wire that lies in the magnetic field is 0.060 m. When the switch is closed, a current of 4.0 A flows through the wire from point P to point Q. Which one of the following statements concerning the effect of the magnetic force on the wire is true?

C) The wire will have no net force acting on it.

D) The wire will be pushed downward, into the plane of the paper.

A) The wire will be pushed to the left.

B) The wire will be pushed to the right.

When a current-carrying wire is placed in a magnetic field, a force is experienced due to the interaction between the magnetic field and the current. In this case, the direction of the force can be determined using the right-hand rule, which indicates that the wire will be pushed downward into the plane of the paper.

Explanation

When a current-carrying wire is placed in a magnetic field, a force is experienced due to the interaction between the magnetic field and the current. In this case, the direction of the force can be determined using the right-hand rule, which indicates that the wire will be pushed downward into the plane of the paper.

27. What is the magnitude of the magnetic force acting on the wire?

A) 0.12 N

D) 67 N

C) 0.48 N

B) 0.24 N

The magnetic force acting on the wire is determined by the equation F = BIL, where B is the magnetic field strength, I is the current passing through the wire, and L is the length of the wire in the magnetic field. Given the values in this scenario, the correct magnitude of the magnetic force is 0.48 N.

Explanation

The magnetic force acting on the wire is determined by the equation F = BIL, where B is the magnetic field strength, I is the current passing through the wire, and L is the length of the wire in the magnetic field. Given the values in this scenario, the correct magnitude of the magnetic force is 0.48 N.

28. A current-carrying, rectangular coil of wire is placed in a magnetic field. The magnitude of the torque on the coil is not dependent upon which one of the following quantities?

A) the magnitude of the current in the loop

C) the length of the sides of the loop

D) the area of the loop

B) the direction of the current in the loop

The torque on a current-carrying loop in a magnetic field is dependent on the magnitude of the current, the length of the sides of the loop, and the area of the loop. The direction of the current in the loop does not affect the magnitude of the torque.

Explanation

The torque on a current-carrying loop in a magnetic field is dependent on the magnitude of the current, the length of the sides of the loop, and the area of the loop. The direction of the current in the loop does not affect the magnitude of the torque.

29. A circular coil consists of 5 loops each of diameter 1.0 m. The coil is placed in an external magnetic field of 0.5 T. When the coil carries a current of 4.0 A, a torque of magnitude 3.93 N-m acts on it. Determine the angle between the normal to the plane of the coil and the direction of the magnetic field.

B) 30°

A) 0°

D) 60°

C) 45°

The correct formula to use in this situation is Torque = NIABsin(theta), where N is the number of loops, I is the current, A is the area of the coil, B is the magnetic field strength, and theta is the angle between the normal to the plane of the coil and the magnetic field direction. By rearranging the formula and plugging in the given values, we can solve for theta and find that the angle is 30 degrees.

Explanation

The correct formula to use in this situation is Torque = NIABsin(theta), where N is the number of loops, I is the current, A is the area of the coil, B is the magnetic field strength, and theta is the angle between the normal to the plane of the coil and the magnetic field direction. By rearranging the formula and plugging in the given values, we can solve for theta and find that the angle is 30 degrees.