Question

A charge $q$ is moving with velocity $(v)$ parallel to a magnetic field $(B)$. Force on charge $q$ due to magnetic field is-
(A).$qvb$
(B).$\dfrac{qv}{B}$
(C).zero
(D).$\dfrac{Bv}{q}$

Answer 1

Hint: When a charge is moving in a magnetic field, a force acts on it which depends on the velocity of the charge, magnetic field and the angle between them. The force is perpendicular to the plane of magnetic field and velocity. Positive and negative charges will have forces with the same magnitude but opposite directions acting on them.

Formulas used:
$\vec{F}=q(\vec{v}\times \vec{B})$
$F=qvB\sin \theta $

Complete answer:
When a charge moves in a magnetic field, the field exerts a force on it. The force acting on charge$q$due to magnetic field $(B)$is given by-
$\vec{F}=q(\vec{v}\times \vec{B})$
$\therefore F=qvB\sin \theta $ --- (1)
Here, $F$is the force acting on the charge
$v$ is the velocity of the charge
$\theta $ is the angle between the direction of magnetic field and motion of the charge
The force vector is the cross product of the field vector and velocity vector. Therefore it is perpendicular to the plane of the motion of charge and magnetic field.
As the charge is moving parallel to the magnetic field so, the angle $\theta $ will be-
$\theta =0$
Substituting the value of $\theta $ in eq (1), we get,
$\begin{align}
  & F=qvB\sin 0 \\
 & \therefore F=0 \\
\end{align}$
Therefore, the force acting on the charge due to the magnetic field is zero as the angle between them is zero.

Hence, the correct option is (C).

Note:
The direction of force acting on the charge is given by the right hand rule which states that if you keep your palm wide open with your thumb at right angle to your hand then if your thumb represents the direction of velocity, your fingers represent the magnetic lines of forces, then the direction in which you push your palm is in the direction of force.