Question

An electron moves with a velocity $1\times {10}^{3}m/s$ in a magnetic field of induction $0.3T$ at an angle ${30}^o$. If $\frac{e}{m}$ of electrons $1.76\times {10}^{11}C/kg$, the radius of the path is nearly

• A. ${10}^{-8}m$
• B. $2\times {10}^{-8}m$
• C. ${10}^{-6}m$
• D. ${10}^{-10}m$

Answer 1

The correct option is A ${10}^{-8}m$

• Radius is given as $r=\frac{mVSin\theta }{qB}$ for electron $q=e$ and it is given that angle between filed and velocity is $\theta ={30}^0$
• put $\frac{q}{m}$$=1.76\times {10}^3\frac{C}{Kg}$ and $B=0.3T$ with $V={10}^3$ we get $r=0.94\times {10}^{-8}meter$ which is nearly equal to ${10}^{-8}meter$
• Option A is correct