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

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. 4.28? What is the force on each case? Which case corresponds to stable equilibrium? 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 cm= 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 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 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.