Question

An electron after being accelerated through a potential difference $100V$ enters a uniform magnetic field of $0.004T$ perpendicular to its direction of motion. Calculate the radius of the path described by the electrons.

Answer 1

Potential difference, $V=100V$

Magnetic field $,B=0.004T$

Force, $F=qV=\frac{m{v}^2}{2}(K.E.)$

$\Rightarrow V=\sqrt{\frac{2qv}{m}}$

for radius of path, $\frac{m{v}^2}{R}=qvB$

$\Rightarrow R=\frac{mv}{qB}=\frac{m}{qB}\times \sqrt{\frac{2qv}{m}}$

$\Rightarrow R=\frac{1}{0.004}\times \sqrt{\frac{9.1\times {10}^{-31}\times 2\times 100}{1.6\times {10}^{-19}}}$

$=843.17\times {10}^{-5}m$

$=8.43mm$

Hence, the answer is $8.43mm.$