Question

A uniform magnetic field of $5000$ gauss is established along the positive z-direction. A rectangular loop of side $20cm$ and $5cm$ carries a current of $10A$ is suspended in this magnetic field. What is the torque on the loop in the different cases shown in the following figures? What is the force in each case? Which case corresponds to stable equilibrium?

Answer 1

Uniform magnetic field of $=5000gause$

A rectangular loop of side $=20cm$ and $5cm$

current$=10A$ suspended in this magnetic field.

$\tau =$ torque on the loop in the different cases.

Solution

$\left(i\right)$ Uniform magnetic field strength,

$B=3000gause=3000\times {10}^{-4}T=0.3T$

Length of the rectangular loop,

$I=70cm$

Width of the rectangular loo,

$b=5cm$

Area of the loop,

$A=l\times b=10\times 5=50c{m}^{2}50\times {10}^{-4}{m}^2$

Current in the loop, $I=12A$

Now, taking the anti- clockwise direction of the current as positive, and vice-versa.

$\left(a\right)$ Torque $\overrightarrow{\tau }=I\overrightarrow{A}\times \overrightarrow{B}$

From the fig, it can be observed that A is normal to the y-z plane and B is directed along the z-axis.

$\tau =12\times (50\times {10}^{-4})\hat{i}\times 0.3\hat{k}=1.8\times {10}^{-2}jN-m$

The torque $1.8\times {10}^{-2}$ is Nm along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.

$\left(ii\right)$ Torque,$\overrightarrow{\tau }=I\overrightarrow{A}\times \overrightarrow{B}=(50\times {10}^{-4}\times 12)\hat{K}\times 0.3\hat{K}=0$

Hence, the torque is zero. The force is also zero.

$\left(iii\right)$ In case, the direction of $I\overrightarrow{A}$ and $\overrightarrow{B}$ is the same and the angle between them is zero, if displaced, they came back to an equilibrium. Hence, its equilibrium is stable.