Question

Electron moves at right angles to a magnetic field of $1.5\times {10}^{-2}$ tesla with speed of $6\times {10}^{7}m/s$. If the specific charge of the electron is $1.7\times {10}^{11}C/kg$. The radius of circular path will be

• A. $3.31cm$
• B. $4.31cm$
• C. $1.31cm$
• D. $2.35cm$

Answer 1

The correct option is D $2.35cm$

$B=1.5\times {10}^{-2}T$,

$\theta ={90}^o,\sin \theta =1,v=6\times {10}^{7}m/s$,

$\frac{e}{m}=1.7\times {10}^{11}C/kg$

$r=\frac{mv}{Be}=\frac{6\times {10}^7}{1.5\times {10}^{-2}\times 1.7\times {10}^{11}}$

$=2.35\times {10}^{-2}m=2.35cm$