Question

A stationary charge is placed in a magnetic field. Will it experience a force? Give reasons to justify your answer.

Answer 1

Hint: A charge which has a velocity with a direction not parallel or anti parallel to the magnetic field, experiences a force known as the Lorentz force. It is absolutely imperative, that the charge is moving and in a direction not aligned with the magnetic field for it to experience a force. The magnitude of this force is given by the Lorentz force law.

Formula used:
$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ (Lorentz force Law)
where $\overrightarrow{F}$ is the Lorentz force on the charged particle, q is the magnitude of charge on the body, $\overrightarrow{v}$ is the velocity of the charged particle and $\overrightarrow{B}$ is the magnetic field vector.
$\left| \overrightarrow{F} \right|=\left| q\overrightarrow{v}\times \overrightarrow{B} \right|=qvB\sin \theta $
where $\theta $ is the angle between the velocity and magnetic field vectors.
The direction of the Lorentz force is perpendicular to the plane containing $\overrightarrow{v}$ and $\overrightarrow{B}$.

Complete Step-by-Step solution:
When a charged particle moves with a velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$, it experiences a force called the Lorentz force. The mathematical expression for this force is given by,
$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ (Lorentz force Law)
q is the magnitude of charge on the particle with correct signs (for positive and negative charges respectively).
The magnitude of this force is given by
$\left| \overrightarrow{F} \right|=\left| q\overrightarrow{v}\times \overrightarrow{B} \right|=qvB\sin \theta $ -----------------(1)
where $\theta $ is the angle between the velocity and magnetic field vectors.
The direction of the Lorentz force is perpendicular to the plane containing $\overrightarrow{v}$ and$\overrightarrow{B}$.
For a stationary charged particle, $v=0$, therefore from equation (1), the force will also be zero.
Hence, a stationary charge will not experience a force in a magnetic field.

Note: Charges in a magnetic field also do not experience a force if they move parallel or anti parallel to the magnetic field$\left( \theta ={{0}^{0}}\text{ or 18}{{\text{0}}^{0}},\sin \theta =0 \right)$. Putting $\theta =0\text{ }or\text{ 18}{{\text{0}}^{0}}$ (for parallel and anti parallel respectively) in equation (1) makes the magnitude of the force equal to zero and hence, the particle does not experience a force.
Another interesting fact to note is that since the force applied by the magnetic field is perpendicular to the plane of the velocity (and hence displacement) of the particle, the work done by it is zero. This is due to the fact that forces acting perpendicular to the displacement of a body do not do work on them.