Physics: Magnetic Forces on Moving Charges Worksheet

Question 1

A proton moves at 2.0 x 10^6 m/s through a magnetic field of 0.30 T. The velocity is perpendicular to the field. Calculate the magnitude of the magnetic force on the proton.

Accepted Answer

The magnetic force is F = qvB sin(theta). Since the motion is perpendicular to the field, F = (1.60 x 10^-19 C)(2.0 x 10^6 m/s)(0.30 T) = 9.6 x 10^-14 N.

Question 2

An electron moves at 5.0 x 10^5 m/s through a 0.80 T magnetic field. The electron's velocity makes a 30° angle with the field. Calculate the magnitude of the magnetic force.

Accepted Answer

The magnetic force is F = |q|vB sin(theta). Substituting gives F = (1.60 x 10^-19 C)(5.0 x 10^5 m/s)(0.80 T)(sin 30°) = 3.2 x 10^-14 N.

Question 3

A positive charge moves east through a magnetic field directed north. Use the right-hand rule to determine the direction of the magnetic force.

Accepted Answer

For a positive charge, point your fingers east and curl them toward north. Your thumb points upward, so the magnetic force is upward.

Question 4

An electron moves north through a magnetic field directed east. Determine the direction of the magnetic force on the electron.

Accepted Answer

For a positive charge, north cross east would point downward. Since an electron is negative, the force is in the opposite direction, so the magnetic force is upward.

Question 5

A particle with charge +2.0 μC moves at 3.0 x 10^4 m/s perpendicular to a 0.50 T magnetic field. Calculate the force magnitude.

Accepted Answer

The force magnitude is F = qvB = (2.0 x 10^-6 C)(3.0 x 10^4 m/s)(0.50 T) = 3.0 x 10^-2 N.

Question 6

A charge of +1.6 x 10^-19 C moves at 2.0 x 10^6 m/s through a 0.25 T magnetic field at an angle of 30° to the field. Calculate the magnetic force magnitude.

Accepted Answer

The force magnitude is F = qvB sin(theta) = (1.6 x 10^-19 C)(2.0 x 10^6 m/s)(0.25 T)(0.5) = 4.0 x 10^-14 N.

Question 7

A charged particle moves exactly parallel to a uniform magnetic field. State the magnetic force on the particle and explain why.

Accepted Answer

The magnetic force is 0 N because F = qvB sin(theta), and theta = 0° for parallel motion. Since sin(0°) = 0, there is no magnetic force.

Question 8

An electron enters a uniform 0.020 T magnetic field with a speed of 4.0 x 10^6 m/s perpendicular to the field. Calculate the radius of its circular path.

Accepted Answer

The radius is r = mv/(|q|B). Using the electron mass, r = (9.11 x 10^-31 kg)(4.0 x 10^6 m/s)/((1.60 x 10^-19 C)(0.020 T)) = 1.1 x 10^-3 m.

Question 9

A proton and an electron move at the same speed perpendicular to the same magnetic field. Compare the radii of their circular paths.

Accepted Answer

The proton has the larger radius. Since r = mv/(|q|B) and the proton and electron have equal charge magnitudes, the radius is proportional to mass. The proton's path radius is about 1836 times the electron's radius.

Question 10

In a velocity selector, an electric field of 2.4 x 10^4 N/C balances a magnetic field of 0.080 T so that charged particles pass through undeflected. Calculate the speed of the particles that pass through.

Accepted Answer

For undeflected motion, electric force equals magnetic force, so qE = qvB. The speed is v = E/B = (2.4 x 10^4 N/C)/(0.080 T) = 3.0 x 10^5 m/s.

Question 11

A singly charged positive ion moves at 2.0 x 10^5 m/s in a 0.40 T magnetic field and follows a circular path with radius 0.12 m. Calculate the mass of the ion.

Accepted Answer

Using r = mv/(qB), solve for mass: m = qBr/v. The mass is m = (1.60 x 10^-19 C)(0.40 T)(0.12 m)/(2.0 x 10^5 m/s) = 3.8 x 10^-26 kg.

Question 12

An alpha particle has charge +3.2 x 10^-19 C and mass 6.64 x 10^-27 kg. It moves at 1.0 x 10^6 m/s perpendicular to a 0.50 T magnetic field. Calculate the force magnitude and acceleration magnitude.

Accepted Answer

The force is F = qvB = (3.2 x 10^-19 C)(1.0 x 10^6 m/s)(0.50 T) = 1.6 x 10^-13 N. The acceleration is a = F/m = (1.6 x 10^-13 N)/(6.64 x 10^-27 kg) = 2.4 x 10^13 m/s^2.

Question 13

A charged particle moves through a uniform magnetic field. Explain why the magnetic force can change the particle's direction but not its speed.

Accepted Answer

The magnetic force is always perpendicular to the particle's velocity. A perpendicular force changes the direction of motion but does no work, so the particle's kinetic energy and speed do not change.

Question 14

An electron moves perpendicular to a magnetic field with speed 3.0 x 10^6 m/s. The radius of its circular path is 0.050 m. Calculate the magnetic field strength.

Accepted Answer

Using r = mv/(|q|B), solve for B: B = mv/(|q|r). Substituting gives B = (9.11 x 10^-31 kg)(3.0 x 10^6 m/s)/((1.60 x 10^-19 C)(0.050 m)) = 3.4 x 10^-4 T.

Question 15

A proton moves in a uniform 0.20 T magnetic field. Its perpendicular velocity component is 4.0 x 10^5 m/s and its parallel velocity component is 3.0 x 10^5 m/s. Calculate the radius of the helical path and the distance advanced along the field in one cycle.

Accepted Answer

The radius is r = mv_perpendicular/(qB) = (1.67 x 10^-27 kg)(4.0 x 10^5 m/s)/((1.60 x 10^-19 C)(0.20 T)) = 2.1 x 10^-2 m. The period is T = 2πm/(qB) = 3.3 x 10^-7 s, so the distance advanced along the field is v_parallel T = (3.0 x 10^5 m/s)(3.3 x 10^-7 s) = 9.8 x 10^-2 m.

Physics: Magnetic Forces on Moving Charges

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