JEE Main 2019 · Physics · Magnetic Effects of Current · medium

A proton enters a uniform magnetic field of 0.1 T with a speed of 2×10^5 m s^-1 in a direction making an angle of 60^° with the field. Take…

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A proton enters a uniform magnetic field of

0.1\ \mathrm{T}

with a speed of

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

in a direction making an angle of

60^\circ

with the field. Take

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

and

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

. The pitch of the helical path is nearly

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Answer: (C) $6.6\ \mathrm{cm}$

The component along the field,

v_\parallel = v\cos60^\circ=1\times10^5\ \mathrm{m\,s^{-1}}

, is unaffected, while the perpendicular component produces circular motion with period

T=\dfrac{2\pi m}{eB}=\dfrac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.1)}\approx6.56\times10^{-7}\ \mathrm{s}

. The pitch is the distance advanced along the field in one period:

p=v_\parallel T\approx 6.56\times10^{-2}\ \mathrm{m}\approx6.6\ \mathrm{cm}

. Using

v\sin60^\circ

instead gives

11.4\ \mathrm{cm}

; the radius of the helix is about

1.8\ \mathrm{cm}

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