Question

An electron moves at right angle to a magnetic field of $15\times {10}^{-2}T$ with a 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 the circular path will be

• A. $2.9cm$
• B. $3.9cm$
• C. $2.35cm$
• D. $2cm$

Answer 1

The correct option is C $2.35cm$

Radius of circular path
$r=\frac{mV}{eB}=\frac{V}{\left(\frac{e}{m}\right)B}$
$r=\frac{6\times {10}^7}{1.7\times {10}^{11}\times 15\times {10}^{-2}}$
$=2.35\times {10}^{-2}m=2.35m$