Question

An electron of kinetic energy of 7.2 × 10*-18 is revolving on circular path in magnetic field 9×10*-5 then radius of its circular path is

Answer 1

Answer: 25 cm

Explanation:

Kinetic energy of moving electron, $K=\frac{1}{2}mv^2$

where, m is the mass and v is the velocity

$\Rightarrow v=\sqrt{\frac{2K}{m}$

Radius of cicular path of an electron in magnetic field is given by,

$R=\frac {mv}{eB}$

where, e is the charge of the electron, m is the mass of the electron, B is the magnitude of the magnetic field and v is the velocity.

$\Rightarrow R= \frac{\sqrt{2Km}}{eB}$

Given,

$K=7.2 \times 10^{-18} kg m^2/s^2$

$B= 9\times 10^{-5}T$

mass of electron, $m=9.1\times 10^{-31} kg$

charge of electron, $e=-1.6\times 10^{-19} C$

Substitute the values,

$R= \frac{\sqrt{2\times 7.2 \times 10^{-18} kg m^2/s^2\times 9.1\times 10^{-31} kg }}{1.6\times 10^{-19} C\times 9 \times 10^{-5}T}=\frac{3.62\times 10^{-24}}{1.44\times 10^{-23}}=0.25 m=25 cm$

Hence, the radius of the circular path is 25 cm.