Question

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

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

Answer 1

Answer: (C)

2.35 cm

Radius of circular path is given by,
$r=\frac{mv}{Be}$

where $B=1.5\times {10}^{-2}T$, $v=6\times {10}^{7}m/s$ and $e/m=1.7\times {10}^{11}C/kg$

$r=\frac{v}{\left(\frac{e}{m}\right)B}=\frac{6\times {10}^7}{1.7\times {10}^{11}\times 1.5\times {10}^{-2}}$
$r=2.35\times {10}^{-2}m=2.35cm$