An electron beam travels with a velocity of $1.6 \times 10^7 \mathrm{~ms}^{-1}$ perpendicularly to magnetic…
An electron beam travels with a velocity of $1.6 \times 10^7 \mathrm{~ms}^{-1}$ perpendicularly to magnetic field of intensity $0.1 \mathrm{~T}$. The radius of the path of the electron beam $\left(m_e=9 \times 10^{-31} \mathrm{~kg}\right.$ )
$9 \times 10^{-5} \mathrm{~m}$
$9 \times 10^{-2} \mathrm{~m}$
$9 \times 10^{-4} \mathrm{~m}$
$9 \times 10^{-3} \mathrm{~m}$
Solution
In a perpendicular magnetic field, the radius of circular path travelled by electron beam is
$\begin{aligned} r & =\frac{m v}{e B} \\ \therefore \quad r & =\frac{9 \times 10^{-31} \times 1.6 \times 10^7}{1.6 \times 10^{-19} \times 0.1} \\ & =9 \times 10^{-4} \mathrm{~m}\end{aligned}$