JEE Main 2018 · Physics · Electrostatics · hard

A proton moving at 2×10^5 m s^-1 enters, perpendicular to its boundary, a region of uniform magnetic field 0.5 T directed perpendicular to the…

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A proton moving at

2\times10^5\ \mathrm{m\,s^{-1}}

enters, perpendicular to its boundary, a region of uniform magnetic field

0.5\ \mathrm{T}

directed perpendicular to the velocity. The region has width

2\ \mathrm{mm}

. Take the proton mass as

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

and charge

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

. The time the proton spends inside the field is nearly

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Answer: (B) $10.5\ \mathrm{ns}$

Radius of the path:

r=\dfrac{mv}{qB}=\dfrac{1.6\times10^{-27}\times2\times10^5}{1.6\times10^{-19}\times0.5}=4\ \mathrm{mm}

. Entering normally to the boundary, the proton has turned through angle

\theta

when it has advanced

r\sin\theta

into the region; it leaves the far side when

r\sin\theta=2\ \mathrm{mm}

, so

\sin\theta=\tfrac12

and

\theta=30^\circ=\pi/6

. The angular speed is

\omega=\dfrac{qB}{m}=5\times10^7\ \mathrm{rad\,s^{-1}}

, so

t=\dfrac{\theta}{\omega}=\dfrac{\pi/6}{5\times10^7}=1.05\times10^{-8}\ \mathrm{s}\approx10.5\ \mathrm{ns}

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