Question

Consider a particle with a charge-to-mass ratio of ?/? = 1 moving in a uniform magnetic field of B = 1 Tesla applied in z-direction. At time t = 0 s, it is located at r = (0, 10, 0) m and its velocity is v = (10, 0, 0) m/s.

(a) Qualitative motion

Draw a diagram of the situation when the proton starts its motion, showing its instantaneous velocity v0, the magnetic field vector B and the direction of the initial force F0 on the proton.

Describe how the velocity of the proton will change as it moves through the magnetic field. Describe changes, if any, in both the magnitude and direction of the velocity.

Describe the path the proton will follow in going through this uniform magnetic field.

Calculate the speed v of the proton.   

[8 marks]

(b) Quantitative motion

Determine the magnetic force acting on the proton initially.

Find the radius of the circular path the proton follows in terms of its charge q, mass m and speed v, and the strength of the magnetic field B. Then, calculate the numerical value of that radius for the proton.

Determine the time required and the angular frequency for one complete revolution.

[6 marks]

(c) Deriving equations of motion

Starting with the magnetic force on the particle and using Newton's second law to write down the differential equations for the components ax , ay and az of the acceleration a acting on the particle.

Describe the motion of the particle in z-direction if the initial velocity component in that direction vz is not zero. Describe how the overall motion of the particle changes in that case .

[6 marks]

Answer 1

a) F₀ will be in the direction of Y axis

The magnitude of velocity will remain unchanged But the direction will gradually change so that the proton undergoes circular motion.

The path followed by the proton will be a circular one with R = mv/qB.

The speed of the proton will remain unchanged and will be equal to initial speed of 10 m/s.

b) The magnetic force is given by F =q*(v X B)

Since v and B are perpendicular to each other, v X B = vB

F = qvB = 10q T =

The magnetic force is given by F = qvB

The centripetal force F = mv²/R balances this force.

qvB = mv²/R

R = mv/qB.

For a proton, R = 1.6726× 10^-27 * 10/(1.602*10^-19 *1) = 1.04406 * 10^-7 m.

Time = distance/velocity = 2*3.14*1.04406* 10^-7 /10 = 6.55669* 10^-8 s.

Angular frequency 9.578* 10⁷ radians per second.

c) F = q(v X B)

using newton's second law, F=ma= m(a_x+a_y+a_z)

Equating this with magnetic force,

The particle will have a uniform motion in the z direction with the velocity component in z direction as well.

The effective motion will be helical.