Question

A point charge of charge $Q$ moving with angular velocity $\omega$ in circle of radius $R$ find $\overrightarrow B$ at centre.

Answer 1

We are given a point charge, angular velocity and radius of the circle in which the point charge is moving. Use the formula of magnetic field due to a point charge. This formula contains terms of velocity, using the relation between the linear velocity and angular velocity, reduce this formula in terms of given values.

Complete step by step answer:
The point charge is having charge of, $Q$
The angular velocity of the point charge is $\omega$
The point charge is moving in a circle of radius $R$
We need to find the magnetic field at the centre:
The magnetic field due to a point charge is given as;
$\overrightarrow B = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{QvR}}{{{R^3}}}$ --equation $1$
Here, $\overrightarrow B$ is the magnetic field.
${\mu _0}$ is a constant known as permeability of free space;
The value of permeability of free space $4\pi \times {10^{ - 7}}\,N{A^{ - 2}}$ and it is connected to the energy stored in a magnetic field.
$v$ is the linear velocity of the point charge;
This formula is in terms of linear velocity but we are given the angular velocity thus, we must convert linear velocity in terms of angular velocity.
The relation between angular velocity and linear velocity is given as:
$v = R\omega$
Substituting the value of linear velocity in equation $1$ , we get
$\overrightarrow B = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{QvR}}{{{R^3}}}$
$\Rightarrow \overrightarrow B = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{Q\left( {R\omega } \right)R}}{{{R^3}}}$
$\Rightarrow \overrightarrow B = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{Q\omega }}{R}$
This is the required magnetic field at the centre of the circle.
This expression is in terms of given values.

Note:
Substitute the linear velocity in terms of angular velocity.
The magnetic field is a vector quantity therefore, it has magnitude as well as direction.
The magnetic field is maximum at the point which is directly perpendicular to the path.