Question

A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig.? What is the force in each case? Which case corresponds to stable equilibrium?

Answer 1

(a)

$A=l\times b=50\times {10}^{-4}$

$\tau =I\overrightarrow{A}\times B=-1.8\times {10}^{-2}\hat{j}Nm$

Force is zero because $\theta =0$

$\tau$ is along negative y-direction.

(b)$A=l\times b=50\times {10}^{-4}$

$\tau =I\overrightarrow{A}\times B=-1.8\times {10}^{-2}\hat{j}Nm$

Force is zero because $\theta =0$

$\tau$ is along negative y-direction.

(c)$A=l\times b=50\times {10}^{-4}$

$\tau =I\overrightarrow{A}\times B=-1.8\times {10}^{-2}\hat{i}Nm$

Force is zero because $\theta =0$

$\tau$ is along negative x-direction.

(d)$A=l\times b=50\times {10}^{-4}$

$|\tau |=IAB=-1.8\times {10}^{-2}Nm$ at angle of ${240}^o$ with x-direction.

Force is zero because $\theta =0$

(e)$A=l\times b=50\times {10}^{-4}$

$\tau =I\overrightarrow{A}\times B$

A is parallel to B. So, $\tau =0$

So, Force is zero.Since the angle is $0^o$, if disturbed it doesn't come back to the original position.

(f)$A=l\times b=50\times {10}^{-4}$

$\tau =I\overrightarrow{A}\times B$

A is opposite to B, i.e., $\theta ={180}^o$.

So, $\tau =0$

So, Force is zero.Since the angle is ${180}^o$, if disturbed it doesn't come back to the original position.