CBSE 2025, Set 3

Question

A rectangular loop of sides $l$ and $b$ carries a current $I$ clockwise. Write the magnetic moment $\vec{m}$ of the loop and show its direction in a diagram.
(ii) The loop is placed in a uniform magnetic field $\vec{B}$ and is free to rotate about an axis which is perpendicular to $\vec{B}$. Prove that the loop experiences no net force, but a torque $\vec{	au} = \vec{m} 	imes \vec{B}$.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Magnetic Moment:**
The magnitude of the magnetic moment is $\vec{m} = I\vec{A}$, where $A = l 	imes b$.
For a clockwise current, using the right-hand grip rule, the direction of the magnetic moment is perpendicularly **into the plane** of the paper.

![Magnetic Moment Direction](https://www.edvaya.com/images/cbse/physics/2025/set3/q25b-i-loop-magnetic-moment-answer.webp)

**(ii) Proof of Zero Net Force and Non-Zero Torque:**
Consider the rectangular loop ABCD placed in a uniform magnetic field $\vec{B}$. Let the normal to the area vector make an angle $	heta$ with $\vec{B}$.

![Torque on Rectangular Loop](https://www.edvaya.com/images/cbse/physics/2025/set3/q25b-ii-loop-torque-forces-answer.webp)

1. **Force on arms parallel to the field component:** The forces on arms BC and DA ($\vec{F}_{BC}$ and $\vec{F}_{DA}$) are equal, opposite, and collinear. They cancel each other out entirely.
2. **Force on perpendicular arms:** The force on arm AB is $\vec{F}_{AB} = I(l)\vec{B}$ (directed outwards) and the force on arm CD is $\vec{F}_{CD} = I(l)\vec{B}$ (directed inwards). 
Since $|F_{AB}| = |F_{CD}| = IlB$, and they act in opposite directions, the **net force on the loop is zero**.

3. **Torque:** However, $\vec{F}_{AB}$ and $\vec{F}_{CD}$ act along different lines of action separated by a perpendicular distance. They form a couple, which exerts a torque on the loop.
The perpendicular distance between the two forces is $b \sin 	heta$.
$$	au = 	ext{Force} 	imes 	ext{perpendicular distance}$$
$$	au = (IlB) 	imes (b \sin 	heta)$$
$$	au = I(lb)B \sin 	heta$$
Since Area $A = lb$ and magnetic moment $m = IA$:
$$	au = mB \sin 	heta$$
In vector form, this is written as:
$$\boxed{\vec{	au} = \vec{m} 	imes \vec{B}}$$

---

**Alternatively:**

![Alternative Torque Diagram](https://www.edvaya.com/images/cbse/physics/2025/set3/q25b-ii-loop-torque-alternative-answer.webp)

If the plane of the current carrying coil makes an angle $\alpha$ with the magnetic field:
* $\vec{F}_{DA} = -\vec{F}_{BC}$ (cancel each other)
* Force on the arm DC is into the plane of the paper: $|F_{DC}| = IbB$
* Force on the arm AB is out of the plane of the paper: $|F_{AB}| = IbB$
* Since forces are equal & opposite, so **net force = 0**.

Both of them form a couple and the magnitude of torque acting on the coil is:
$	au =$ either force $	imes$ perpendicular distance between the two forces.
$$	au = IbB 	imes l \sin 	heta$$
$$\vec{	au} = I\vec{A} 	imes \vec{B}$$
$$\boxed{\vec{	au} = \vec{m} 	imes \vec{B}}$$

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