Question

The figure shows a region of length ' $l$' with a uniform magnetic field of $0.3T$ in it and a proton entering the region with velocity $4\times 10^{5}m{s}^{-1}$ making an angle $60^{\circ }$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, ${}^{\prime }{l}^{\prime }$ is close to (mass of proton $=1.67\times 10^{-27}kg,$ charge of the proton $=1.6\times 10^{-19}C)$

• A. $0.11m$
• B. $0.22m$
• C. $0.44m$
• D. $0.88m$

Answer 1

Answer: (C)

$T=\frac{2\pi m}{qB}$
total time $t=10T$

Kinematics
$\ell =\frac{V}{2}t$
$\ell =\frac{V}{2}10\times \frac{2\pi m}{qB}$
$=4\times 10^5\times 10\times \frac{3.14\times 1.67\times 10^{-27}}{1.6\times 10^{-19}\times 0.3}$
$=0.439$