Question

A 10 eV electron is circulating in a plane at right angles to a uniform field of magnetic induction ${10}^{-4}Wb/m^2$(=1.0 gauss), the orbital radius of electron is

• A. 11 cm
• B. 18 cm
• C. 16 cm
• D. 12 cm

Answer 1

The correct option is A 11 cm

$\text{Given}$: Kinetic energy of electron ( $\frac{1}{2}m{v}^2$ ) = 10 eV = 1$0\times (1.6\times {10}^{-19})$ J

Magnetic induction (B) = ${10}^{-4}$ Wb/$m^2$

$\text{To find}$ : Orbital radius r

$\text{Solution}$:

We know mass of electron= $9.1\times {10}^{-31}$Kg

Therefore, $\frac{1}{2}(9.1\times {10}^{-31})v^2=10\times (1.6\times {10}^{-19})$

Or, $v^2$= $3.52\times {10}^{12}$

Or, v = $1.876\times {10}^6$ m

Now centripetal force = $\frac{m{v}^2}{r}$ = Bev

Therefore, r= mv/Be = $\frac{9.1\times {10}^{-31}\times 1.876\times {10}^6}{{10}^{-4}\times 1.6\times {10}^{-19}}$

Hence r= $11\times {10}^{-2}$ m = 11 cm

$\text{Hence A is the correct option.}$