Question:
The figure shows a region of length ' $l$ ' with a uniform magnetic field of $0.3 \mathrm{~T}$ in it and a proton entering the region with velocity $4 \times 10^{5} \mathrm{~ms}^{-1}$ 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
$=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton
$\left.=1.6 \times 10^{-19} \mathrm{C}\right)$
Correct Option: , 3
Solution:
$T=\frac{2 \pi m}{q B}$
total time $\mathrm{t}=10 \mathrm{~T}$
Kinematics
$\ell=\frac{\mathrm{V}}{2} \mathrm{t}$
$\ell=\frac{V}{2} 10 \times \frac{2 \pi m}{q B}$
$=4 \times 10^{5} \times 10 \times \frac{3.14 \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.3}$
$=0.439$