Question

Refer to the figure, rectangle $PQRS$ represents the cross section of a uniform magnetic field region of $0.20T$. An electron is projected at a speed of $v=2.0\times {10}^{6}m/s$ into the region at an angle of ${30}^{\circ }$ to the direction of the magnetic field. The length of the magnetic field region is $0.01m$. Find the number of revolutions made by the electron before it leaves the magnetic field region.

Answer 1

Velocity component to field is $V_1=V\cos 30°=\frac{\sqrt{3}V}{2}$

$V_1=\frac{\sqrt{3}V}{2}\times 2\times 10=\sqrt{3}\times {10}^6$

Time taken out of field=$\frac{0.01}{V_1}$

$t=\frac{0.01}{\sqrt{3}\times {10}^6}=\frac{1}{\sqrt{3}}\times {10}^{-8}sec$

Velocity component parallel to field is $V_2=V\sin 30°$

$V_2=1\times {10}^6$

$r$=radius of circular motion of helical path.

$\frac{m{V_2}^2}{r}=\overrightarrow{q}\overrightarrow{V_2}\overrightarrow{B}$ =Force of electron

$q$=charge on electron

$m$=mass of electron

$r=\frac{m{V}_2}{qB}$

Time taken to cover one revolution is $t_1$

$t_1=\frac{2\pi r}{\sqrt{2}}=\frac{2\pi m}{qB}$

$t_1=\frac{2\times 3.14\times 9.1\times {10}^{-31}}{1.6\times {10}^{-19}\times 0.2}=178.5\times {10}^{-12}$

Number of revolution$\left(n\right)=\frac{t}{t_1}=\frac{\frac{1}{\sqrt{3}}\times {10}^{-8}}{178.5\times {10}^{-12}}=32.89\approx 33$ revolutions.