Question

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

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

Answer 1

Answer: (C)

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

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