A proton (mass 1.67×10^-27 kg, charge…

A proton (mass 1.67×10^-27 kg, charge 1.6×10^-19 C) is released from rest at the origin in uniform fields vec E=2.0×10^3 V m^-1 hat x and vec B=0.10…

One or more options correct

A proton (mass

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

, charge

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

) is released from rest at the origin in uniform fields

\vec E=2.0\times10^{3}\ \mathrm{V\,m^{-1}}\ \hat x

and

\vec B=0.10\ \mathrm{T}\ \hat z

. Gravity is negligible. Which of the following statements is/are correct?

A. The maximum speed attained by the proton is $4.0\times10^{4}\ \mathrm{m\,s^{-1}}$
B. The maximum displacement of the proton along the $x$ -axis is about $2.1\ \mathrm{mm}$
C. In one cyclotron period the proton drifts by about $2.6\ \mathrm{cm}$ along the $-y$ direction
D. The maximum kinetic energy of the proton equals $4\times\tfrac12m\left(E/B\right)^2$

Show answer and solution

Answer: A, D

In crossed fields the motion is a cycloid: a uniform drift of speed

v_d=E/B=2.0\times10^{4}\ \mathrm{m\,s^{-1}}

along

\hat E\times\hat B=-\hat y

plus circular motion; starting from rest the circular speed equals

v_d

. The maximum speed is therefore

2v_d=4.0\times10^{4}\ \mathrm{m\,s^{-1}}

, so (A) is correct. The maximum kinetic energy is

\tfrac12m(2v_d)^2=4\times\tfrac12mv_d^2

, so (D) is correct. The generating circle has radius

r=mv_d/(eB)=mE/(eB^2)\approx2.09\ \mathrm{mm}

and the farthest displacement along

\hat x

2r\approx4.2\ \mathrm{mm}

, not

2.1\ \mathrm{mm}

; (B) is wrong. The period is

2\pi m/(eB)\approx6.56\times10^{-7}\ \mathrm{s}

, so the drift per cycle is

v_dT\approx1.3\ \mathrm{cm}

along

-\hat y

, not

2.6\ \mathrm{cm}

; (C) is wrong. Answer: A and D.

Original question written in the JEE Advanced 2018 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake