A proton of mass 1.67×10^-27 kg and…

A proton of mass 1.67×10^-27 kg and charge 1.6×10^-19 C enters a uniform magnetic field of 0.20 T with speed 1.0×10^6 m s^-1, its velocity making an…

One or more options correct

A proton of mass

1.67\times10^{-27}\ \mathrm{kg}

and charge

1.6\times10^{-19}\ \mathrm{C}

enters a uniform magnetic field of

0.20\ \mathrm{T}

with speed

1.0\times10^{6}\ \mathrm{m\,s^{-1}}

, its velocity making an angle of

30^\circ

with the field direction. Which of the following statements is/are correct?

A. The radius of the helical path is $2.61\ \mathrm{cm}$ .
B. The pitch of the helix is $2.84\ \mathrm{cm}$ .
C. The time taken for one turn of the helix increases if the speed of the proton is doubled, the angle remaining the same.
D. The magnetic force does work $qvB\sin30^\circ\times 2\pi r$ on the proton in one turn of the helix.

Show answer and solution

Answer: A

Resolve the velocity:

v_\parallel=v\cos30^\circ

is unaffected by the field and

v_\perp=v\sin30^\circ=5\times10^5\ \mathrm{m\,s^{-1}}

produces circular motion. The radius is

r=\dfrac{mv_\perp}{qB}=\dfrac{1.67\times10^{-27}\times5\times10^5}{1.6\times10^{-19}\times0.2}=2.61\ \mathrm{cm}

(A true). The pitch is the distance moved along the field during one period,

p=v_\parallel T=v\cos30^\circ\cdot\dfrac{2\pi m}{qB}\approx28.4\ \mathrm{cm}

, ten times larger than the quoted

2.84\ \mathrm{cm}

(B false). The period

T=2\pi m/qB

does not depend on the speed, so it stays the same (C false). The magnetic force is always perpendicular to the velocity, so it does no work in any path (D false); only the direction of the velocity changes.

Report a mistake

← Previous Magnetic Effects of Current question Next Magnetic Effects of Current question →

Original question written in the JEE Advanced 2016 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake