Question

An electron having a kinetic energy of 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.

Answer 1

$KE=100eV=1.6\times {10}^{-17}J$

$\frac{1}{2}\times 9.1\times {10}^{-31}\times V^2=1.6\times {10}^{-17}J$

$\Rightarrow V^2=0.35\times {10}^{14}$

$or,V=0.591\times {10}^{7}m/s$

Now, $r=\frac{mV}{qB}\Rightarrow B=\frac{mV}{qr}$

$B=3.3613\times {10}^{-4}T$

$=3.4\times {10}^{-4}T$

No. of cycles per second

$=\frac{1}{T}=0.0951\times {10}^8$

$=9.51\times {10}^8$

Putting $B=3.361\times {10}^{-4}T$

We get $f=9.4\times {10}^8$