Question

A charged particle is moving in a magnetic field of strength B perpendicular to the direction of the field. If q and m denote the charge and mass of the particle respectively. Then the frequency of rotation of the particle is

Answer 1

The force on a charged particle due to an electric field is directed parallel to the electric field vector in the case of a positive charge, and anti-parallel in the case of a negative charge. It does not depend on the velocity of the particle.

In contrast, the magnetic force on a charge particle is orthogonal to the magnetic field vector, and depends on the velocity of the particle. The right hand rule can be used to determine the direction of the force.

An electric field may do work on a charged particle, while a magnetic field does no work.

The Lorentz force is the combination of the electric and magnetic force, which are often considered together for practical applications.

Electric field lines are generated on positive charges and terminate on negative ones. The field lines of an isolated charge are directly radially outward. The electric field is tangent to these lines.

Magnetic field lines, in the case of a magnet, are generated at the north pole and terminate on a south pole. Magnetic poles do not exist in isolation. Like in the case of electric field lines, the magnetic field is tangent to the field lines. Charged particles will spiral around these field lines.

Answer 2

Given:

Magnetic field strength $=B$

Charge on the particle $=q$

Mass of the particle $=m$

To Find: Frequency of rotation of the particle

Solution:

When a particle moves in a magnetic field, a magnetic force acts on the particle.

Suppose the speed of the moving charged particle having charge $q$ is $v$ and it enters the magnetic field $B$ at an angle $\[\theta \]$.

Then, according to the Lorentz force, the force acting on the particle moving in a magnetic field is given as:

$\[F = qvB\sin \theta \]$              ... (i)

As the particle is moving perpendicular to the magnetic field, therefore, $\[\theta = 90^\circ \]$.

Now, the force becomes

$\[F = qvB\]$                       ... (ii)

Since the particle is rotating in a magnetic field, it performs a circular motion. Therefore, a centripetal force acts on it. Thus,

The centripetal force is given by

$\[F = \frac{{m{v^2}}}{r}\]$                       ... (iii)

where, $r$ is the radius of the circle.

Here, the centripetal force is just a magnetic force. Thus,

Equating the above two equations (ii) and (iii), we have,

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

$\[r = \frac{{mv}}{{qB}}\]$                         ... (iv)

If the time period of the circular motion of the charge is $T$, then,

$\[v = \frac{{2\pi r}}{T}\]$                        ... (v)

Substituting the value of $r$ from equation (iv) in the above equation (v), we have,

$\[\begin{gathered} v = \frac{{2\pi mv}}{{qBT}} \hfill \\ T = \frac{{2\pi m}}{{qB}} \hfill \\ \end{gathered} \]$

Thus, the frequency is given as:

$\[f = \frac{1}{T} = \frac{{qB}}{{2\pi m}}\]$

Hence, the frequency of rotation of the particle is $\[f = \frac{{qB}}{{2\pi m}}\]$.

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