Question

A proton is moving with a uniform velocity of $10^{6}m{s}^{-1}$ along the $Y-axis$, under the joint action of a magnetic field along $Z-axis$ and an electric field of magnitude $2\times 10^{4}V{m}^{-1}$ along the negative $X-axis$. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly (given : $\frac{e}{m}$ ratio for proton $\approx 10^{8}Ck{g}^{-1}$)

• A. 0.5 m
• B. 0.2 m
• C. 0.1 m
• D. 0.05 m

Answer 1

Answer: (A)

Velocity of proton $=10^{6}m/s$ along $y$-direction

Electric field $=2\times 10^{4}V/m$
$\frac{e}{m}$ for proton $=10^{-8}C/kg$
The magnetic field,
$B=\frac{E}{v}\left(\because q_{p}vB=q_{p}E\right)$
$=\frac{2\times 10^4}{10^6}=2\times 10^{-2}T$
The radius of circular path
$\gamma =\frac{mv}{q_p\cdot B}=\frac{10^{-8}\times 10^6}{2\times 10^{-2}}$
$=\frac{1}{2}=0.5m$