Figure shows a region of length 'l' with a uniform magnetic field of 0.3, T in it and a prot

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4.Moving Charges and Magnetism

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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, $l$ is close to....... $m$

(mass of proton $=1.67 \times 10^{-27} \,kg ,$ charge of the proton $\left.=1.6 \times 10^{-19}\, C \right)$

A

$0.11$

B

$0.22$

C

$0.44$

D

$0.88$

(JEE MAIN-2020)

Solution

Sol. $\quad T =\frac{2 \pi m }{ qB }$

total time $t=10 T$

Kinematics

$\ell=\frac{ V }{2} t$

$\ell=\frac{V}{2} 10 \times \frac{2 \pi m}{q B}$

$=4 \times 10^{5} \times 10 \times \frac{3.14 \times 1.67 \times 1 0^{-27}}{1.6 \times 10^{-19} \times 0.3}$

$=0.439$

Standard 12

Physics

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