Physics: Magnetic Forces on Moving Charges
Calculating force magnitude, direction, and charged particle motion in magnetic fields
Physics: Magnetic Forces on Moving Charges
Calculating force magnitude, direction, and charged particle motion in magnetic fields
Physics - Grade 9-12
- 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.
For perpendicular velocity and magnetic field, sin(90°) = 1.
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. - 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.
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. - 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.
Use the direction of v cross B for a positive charge.
For a positive charge, point your fingers east and curl them toward north. Your thumb points upward, so the magnetic force is upward. - 4
An electron moves north through a magnetic field directed east. Determine the direction of the magnetic force on the electron.
Find the right-hand-rule direction first, then reverse it for a negative charge.
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. - 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.
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. - 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.
Only the component of velocity perpendicular to the magnetic field contributes to magnetic force.
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. - 7
A charged particle moves exactly parallel to a uniform magnetic field. State the magnetic force on the particle and explain why.
A magnetic field only exerts force on the part of the velocity that is perpendicular to the field.
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. - 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.
For circular motion in a magnetic field, set qvB equal to mv^2/r.
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. - 9
A proton and an electron move at the same speed perpendicular to the same magnetic field. Compare the radii of their circular paths.
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. - 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.
The charge cancels out when electric and magnetic forces balance.
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. - 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.
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. - 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.
Find the magnetic force first, then use Newton's second law.
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. - 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.
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. - 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.
Rearrange the circular motion equation before substituting values.
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. - 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.
The perpendicular component sets the circular radius, and the parallel component sets the pitch of the helix.
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.