Question

A 10 ev electron is circulating in a plane at right angles to a uniform field at amgnetic induction 10^-4 wb/m^2. the orbital radius will be

Answer 1

A 10 eV electron with kinetic energy $k =\frac{1}{2} m v^{2}$ circles in a plane perpendicular to uniform magnetic field.

Therefore we can write,$eV=\frac{1}{2} mv^{2}$

or

$v=\sqrt{\frac{2e V}{m} }$

Here, m is the mass of electron and v is its speed.

As,  $e=1.6\times10^{-19} C$,$m=9.1\times10^{-31} kg$.

Substituting these values in above formula we get

$v=\sqrt{\frac{2\times1.6\times10^{-19}\times10 }{9.1\times10^{-31} } }$

$v=1.89\times10^{6} m/s$

The magnetic force supplies the centripetal force, therefore

$qvB=\frac{mv^{2} }{r}$

or

$r=\frac{mv}{qB}$

Given,   $B=10^{-4}  T$

Thus, substituting the all the values in above formula we can calculate the orbital radius,

$r=\frac{9.1\times10^{-31}\times1.89\times10^{6}  }{1.6\times10^{-19}\times10^{-4}  }$

$r=0.106 m \simeq 11 cm$