Question

A particle of mass m with charge q moving with a uniform speed v normal to a uniform magnetic field B. Derive the expression for (i) radius of the circular path, (ii)time-period of revolution, and (ii) kinetic energy of the particle.?

Answer 1

Hint: We all know that when a charged particle moves in a magnetic field, a magnetic force acts on a particle, and the particle deflects occasional, and the magnitude of the magnetic force depends on the charge of the particle, speed of the particle and the Intensity of magnetic field.

Complete step by step solution:
The centripetal force balances the magnetic force acting on a moving particle perpendicular to its motion. We can write the equilibrium formula as,
$F = {F_c}$ …… (I)
Here, $F$ is the force on a charged particle ${F_c}$ is the centripetal force. We know that F is the force on a charged particle in a magnetic field and is given by,
$F = qvB$
Here q is the charge, v is the velocity of the particle, and B is the magnetic field.
The centripetal force is given by,
${F_c} = \dfrac{{m{v^2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.
We will now substitute ${F_c} = \dfrac{{m{v^2}}}{r}$ $F = qvB$ in equation (I) to find the value of r.
$\begin{array}{l}
qvB = \dfrac{{m{v^2}}}{r}\\
r = \dfrac{{mv}}{{qB}}
\end{array}$
We know that the period of revolution is given by,
\[T = \dfrac{{2\pi r}}{v}\] …… (II)
Here r is the radius, and v is the velocity of the particle.
We will now substitute \[r = \dfrac{{mv}}{{qB}}\] in equation (II) to find the value of T.
\[\begin{array}{l}
T = \dfrac{{2\pi }}{v}\left( {\dfrac{{mv}}{{qB}}} \right)\\
T = \dfrac{{2\pi m}}{{qB}}
\end{array}\]

The kinetic energy $E$ on the particle will remain equal to $\dfrac{{m{v^2}}}{2}$ as the particle is under the same field moving with velocity v.

Note: The force on a particle acts when there is some relative motion between the particle and magnetic field. Suppose the magnetic field and the particle are moving at the same speed, then the force acting on the particle is zero.