JEE Main 2016 · Physics · Electrostatics · medium

A proton enters a uniform magnetic field with its velocity making an angle of 60^° with the field direction and describes a helical path. The ratio…

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A proton enters a uniform magnetic field with its velocity making an angle of

60^\circ

with the field direction and describes a helical path. The ratio of the pitch of the helix to the radius of the helix is

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Answer: (A) $\dfrac{2\pi}{\sqrt3}$

Resolve the velocity:

v_\perp=v\sin60^\circ

gives circular motion of radius

r=\dfrac{mv\sin60^\circ}{qB}

and time period

T=\dfrac{2\pi m}{qB}

, while

v_\parallel=v\cos60^\circ

is unchanged by the field. In one period the proton advances along the field by the pitch

p=v_\parallel T=\dfrac{2\pi mv\cos60^\circ}{qB}

. Dividing,

\dfrac pr=2\pi\cot60^\circ=\dfrac{2\pi}{\sqrt3}\approx3.63

. Using

\tan60^\circ

instead of

\cot60^\circ

gives the wrong option

2\pi\sqrt3

, so keep track of which component is along the field.

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