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 of the electron is $1.7\times 10^{11}C/kg$. The radius of the circular nath will he

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

Answer 1

Answer: (C)

The formula for radius of circular path is
$r=\frac{mv}{eB}=\frac{v}{\left(\frac{e}{m}\right)B}...(1)$
Given: $e/m$ of electron
$=1.7\times 10^{11}C/kg$ and $v=6\times 10^{7}m/s$
$B=1.5\times 10^{-2}T$
So, $r=\frac{6\times 10^7}{1.7\times 10^{11}\times 1.5\times 10^{-2}}$
$=2.35\times 10^{-2}m$
$=2.35cm$