Question

An electron is accelerated through
a potential difference of 2000 V and then it enters a
uniform magnetic field of 0.02 Tesla in a direction
perpendicular to it. Find the radius of the path of the
electron in the magnetic field.
​

Answer 1

The radius of the path of the electron in the magnetic field is $7.54\cdot 10^{-3} m$

Explanation:

First of all, we have to find the initial speed of the electron. When it is accelerated by the potential difference, the electric potential energy is converted into kinetic energy, therefore:

$q\Delta V = \frac{1}{2}mv^2$

where

$q=1.6\cdot 10^{-19} C$ is the electron charge

$\Delta V=2000 V$ is the potential difference

$m=9.11\cdot 10^{-31}kg$ is the electron mass

v is the final speed of the electron

Solving for v,

$v=\sqrt{\frac{2q\Delta V}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})(2000)}{9.11\cdot 10^{-31}}}=2.65\cdot 10^7 m/s$

When a charged particle is moving in a magnetic field, it experiences a force (perpendicular to the direction of motion). The force acting on the particle is given by (for a magnetic field perpendicular to the motion of the particle)

$F=qvB$

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

This force acts as a centripetal force, so the particle acquires a uniform circular motion. We can write therefore

$qvB=m\frac{v^2}{r}$

where m is the mass of the particle and r the radius of the trajectory.

Solving for r,

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

Here we have

B = 0.02 T

And using m, v and q of the electron, we find:

$r=\frac{(9.11\cdot 10^{-31})(2.65\cdot 10^7)}{(1.6\cdot 10^{-19})(0.02)}=7.54\cdot 10^{-3} m$

Learn more about magnetic fields:

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