Question

A proton is moving with a uniform velocity of $10^6m\,s^{-1}$ along the $Y-axis$, under the joint action of a magnetic field along $Z-axis$ and an electric field of magnitude $2×10^4 \, Vm^{-1}$ 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 $\approx 10^{8} Ckg^{-1}$)

• 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,
$B =\frac{E}{v} \quad\left(\because q_{p} v B=q_{p} E\right)$
$=\frac{2 \times 10^{4}}{10^{6}}=2 \times 10^{-2} T$
The radius of circular path
$\gamma =\frac{m v}{q_{p} \cdot B}=\frac{10^{-8} \times 10^{6}}{2 \times 10^{-2}} $
$=\frac{1}{2}=0.5 \,m$