Question

A moving charge can produce a magnetic field. How does a current loop behaves like a magnetic dipole?

Answer 1

When an electric current flows in a closed loop of wire, placed in a uniform magnetic field, the magnetic forces produce a torque which tends to rotate the loop so that area of the loop is perpendicular to the direction of the magnetic field.

Consider a rectangular coil $PQRS$ placed in an external magnetic field as shown in diagram (a). Let ${}^{\prime }{I}^{\prime }$ be the current flowing through the coil. Each part of the coil experiences one Lorentz. Each part of the forces is$\overrightarrow{F_1},\overrightarrow{F_2},\overrightarrow{F_3}$ and $\overrightarrow{F_4}$ as shown. The forces are equal in magnitude but act in opposite directions along the same straight line. Hence they cancel out

The force $\overrightarrow{F_1}=I(\overrightarrow{PQ}\times \overrightarrow{B})$

$F_1=IlB$ when $\theta ={90}^0$ $\overrightarrow{F_1}$ acts in direction perpendicular to the plane of paper.

Similarly, $\overrightarrow{F_3}=I(\overrightarrow{RS}\times \overrightarrow{B}$

$F_3=IB$ when $\theta ={90}^0$

The two forces constitute a couple and so rotates the coil in the anticlockwise. The torque is given by

$\tau =\text{force x arm of couple}$

$\tau =Fbcos\theta$

$\tau =IlBbcos\theta$

$\tau =IABcos\theta$

The area vector $A$ makes an angle $\alpha$ with the $\overrightarrow{B}$ so $\theta +\alpha ={90}^0$

$cos\theta =cos(90-\alpha )=sin\alpha$

$\therefore \tau =NIABsin\alpha$

$\tau =P_{m}Bsin\alpha$

$\overrightarrow{\tau }=\overrightarrow{P_m}\times \overrightarrow{B}$ where $P_m=NIA$ called the magnetic dipole moment of the loop.