An electron having kinetic energy of $100 \mathrm{eV}$ circulates in a path of radius $10 \mathrm{~cm}$ in a…
An electron having kinetic energy of $100 \mathrm{eV}$ circulates in a path of radius $10 \mathrm{~cm}$ in a magnetic field. The magnitude of magnetic field $|\mathbf{B}|$ is approximately [Mass of electron $=0.5 \mathrm{MeV} \mathrm{c}^{-2}$, where $\mathrm{c}$ is the velocity of light].
$3.3 \times 10^{-4} \mathrm{~T}$
$2.6 \times 10^{-4} \mathrm{~T}$
$1.70 \times 10^{-4} \mathrm{~T}$
$4.3 \times 10^{-4} \mathrm{~T}$
Solution
Kinetic energy of electron,
$\begin{aligned} K & =100 \mathrm{eV} \\ & =100 \times 1.5 \times 10^{-19} \mathrm{~J}=1.6 \times 10^{-17} \mathrm{~J}\end{aligned}$
Radius of circular path, $r=10 \mathrm{~cm}=0.1 \mathrm{~m}$ Mass of electron, $m=0.5 \mathrm{MeVc}^{-2}$
$=\frac{0.5 \mathrm{MeV}}{c^2}=\frac{0.5 \times 10^6 \times 1.6 \times 10^{-19}}{3 \times 10^8 \times 3 \times 10^8} \mathrm{~kg}$
$=8.89 \times 10^{-31} \mathrm{~kg}$
$\because$ Radius of circular path of electron in magnetic field $B$ in terms of kinetic energy $(K)$ is given as
$r=\frac{\sqrt{2 m K}}{B q}$
$\Rightarrow \quad B=\frac{\sqrt{2 m K}}{r q}=\frac{\sqrt{2 \times 8.89 \times 10^{-31} \times 1.6 \times 10^{-17}}}{0.1 \times 1.6 \times 10^{-19}}$
$=3.3 \times 10^{-4} \mathrm{~T}$