A uniform magnetic field of 3000 G is established along the positive z-direction.

Solution:

Magnetic field strength, B = 3000 G = 3000 × 10−4 T = 0.3 T

Length of the rectangular loop, l = 10 cm

Width of the rectangular loop, b = 5 cm

Area of the loop,

A = l × b = 10 × 5 = 50 cm2 = 50 × 10−4 m2

Current in the loop, I = 12 A

Now, taking the anti-clockwise direction of the current as positive and vise-versa:

(a) Torque, $\vec{\tau}=I \vec{A} \times \vec{B}$

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

$\therefore \tau=12 \times\left(50 \times 10^{-4}\right) \hat{i} \times 0.3 \hat{k}$

$=-1.8 \times 10^{-2} \hat{j} \mathrm{Nm}$

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

(b) This case is similar to case (a). Hence, the answer is the same as (a).

(c) Torque $\tau=I \vec{A} \times \vec{B}$

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

$\therefore \tau=-12 \times\left(50 \times 10^{-4}\right) \hat{j} \times 0.3 \hat{k}$

$=-1.8 \times 10^{-2} \hat{i} \mathrm{Nm}$

The torque is $1.8 \times 10^{-2} \mathrm{~N} \mathrm{~m}$ along the negative $x$ direction and the force is zero.

(d) Magnitude of torque is given as:

$|\tau|=I A B$

$=12 \times 50 \times 10^{-4} \times 0.3$

$=1.8 \times 10^{-2} \mathrm{~N} \mathrm{~m}$

Torque is $1.8 \times 10^{-2} \mathrm{~N} \mathrm{~m}$ at an angle of $240^{\circ}$ with positive $x$ direction. The force is zero.

(e) Torque $\tau=I \vec{A} \times \vec{B}$

$=\left(50 \times 10^{-4} \times 12\right) \hat{k} \times 0.3 \hat{k}$

= 0

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

(f) Torque $\tau=I \vec{A} \times \vec{B}$

$=\left(50 \times 10^{-4} \times 12\right) \hat{k} \times 0.3 \hat{k}$

= 0

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

In case (e), the direction of $I \vec{A}$ and $\vec{B}$ is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.

Whereas, in case (f), the direction of $I \vec{A}$ and $\vec{B}$ is opposite. The angle between them is $180^{\circ}$. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.