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)

2.35 cm

Answer 2

The formula for radius of circular path is
$r=\frac{m v}{e B}=\frac{v}{\left(\frac{e}{m}\right) B}\,\,\,... (1)$
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, $r=\frac{6 \times 10^{7}}{1.7 \times 10^{11} \times 1.5 \times 10^{-2}}$
$=2.35 \times 10^{-2} m$
$=2.35 \,cm$