Question

(a) Write the expression for the force $\overrightarrow{F}$, acting on a charged particle of charge ${}^{\prime }{q}^{\prime }$, moving with a velocity $\overrightarrow{v}$ in the presence of both electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$. Obtain the condition under which the particle moves undeflected through the fields.
(b) A rectangular loop of size $l\times b$ carrying a steady current $I$ is placed in a uniform magnetic field $\overrightarrow{B}$. Prove that the torque $\overrightarrow{\tau }$ acting on the loop is given by $\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}$ where $\overrightarrow{m}$ is the magnetic moment of the loop.

Answer 1

(a)

Force acting on a charge 'q' moving with a velocity $\overrightarrow{v}$ in the presence of both electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$.

$F=q\overrightarrow{E}+q(v\times \overrightarrow{B}$

Consider a region in which magnetic field, electric field and the velocity of the charge particle are perpendicular to each others. To move charge particle undeflected, the net force on the charge particle must be zero.

$q\overrightarrow{E}=qv\overrightarrow{B}$

$v=\frac{\overrightarrow{E}}{\overrightarrow{B}}$

The direction of the magnetic and electric field are equal and opposite direction. So the magnitude are in such a way they cancel out each other to give a net force of zero. So that the charge particle does not deflect.

(b)

Plane of the loop i at an angle with the direction of the magnetic field. Let $\theta$ be the angle between normal and the field. The forces on BC and DA are equal and opposite and they cancel each other.

Force on AB is $F_1$ and the force on CD is $F_2$

$F_1=F_2=IbB$

Magnitude of the torque is shown in the figure

$\tau =F_1\frac{a}{2}sin\theta +F_2\frac{a}{2}sin\theta$

$=IbB\frac{a}{2}sin\theta +IbB\frac{a}{2}sin\theta$

$=I\left(ab\right)Bsin\theta$

$=IABsin\theta$ ; where $A=ab$

If there are 'n' such turns then there torque will be $nIABsin\theta$

Magnetic moment of the current, $m=IA$

$\overrightarrow{\tau }$= $\overrightarrow{m}\times \overrightarrow{B}$