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 $={10}^{8}C{kg}^{-1}$

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

Answer 1

Answer: (A)

$0.5m$

Initially $F_E=F_m$
$\Rightarrow B=E/V=\left(2/100\right)T$
Now when $E$ switched off
$R=\frac{mv}{qB}=\frac{{10}^6\times 100}{{10}^8\times 2}=0.5m$