JEE Advanced 2018 Paper 1 · Physics · Magnetic Effects of Current · medium

A beam of protons, each of kinetic energy 1.0 MeV, enters a region of uniform magnetic field 0.50 T perpendicular to the beam. The region is bounded…

Numerical answer

A beam of protons, each of kinetic energy

1.0\ \mathrm{MeV}

, enters a region of uniform magnetic field

0.50\ \mathrm{T}

perpendicular to the beam. The region is bounded by two parallel planes a distance

0.10\ \mathrm{m}

apart, and the protons enter normally to the first plane. Use

m_p=1.673\times10^{-27}\ \mathrm{kg}

and

e=1.602\times10^{-19}\ \mathrm{C}

. Find the time, in nanoseconds, for which a proton remains inside the field region. Give the answer correct to two decimal places.

Show answer and solution

Answer: 7.38

The speed follows from

K=\tfrac12m_pv^2

, giving

v=1.384\times10^{7}\ \mathrm{m\,s^{-1}}

. The orbit radius is

r=\dfrac{m_pv}{eB}=0.289\ \mathrm{m}

, which exceeds the width

d=0.10\ \mathrm{m}

, so the proton emerges from the far plane. Because it enters normally, the angle through which its velocity turns satisfies

\sin\theta=d/r=0.346

, so

\theta=20.24^\circ=0.3533\ \mathrm{rad}

. The time spent is the fraction of a cyclotron motion with angular frequency

\omega=eB/m_p

, so

t=\dfrac{\theta m_p}{eB}=0.3533\times\dfrac{1.673\times10^{-27}}{1.602\times10^{-19}\times0.5}=7.38\times10^{-9}\ \mathrm{s}

, i.e.

7.38\ \mathrm{ns}

. Note that

t

is not

d/v

(

7.2\ \mathrm{ns}

), since the path is curved.

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