Q.(a) A circular coil of $30$ turns and radius $8.0\ \text{cm}$ carrying a current of $6.0\ \text{A}$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0\ \text{T}$. The field lines make an angle of $60^\circ$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Concept: Magnetic Force Balance

A current-carrying coil in a magnetic field experiences a torque that tries to align its normal with the field. To keep it stationary, an equal and opposite external torque must be applied.

Reasoning

(a) The torque on a planar coil is given by:

where $N = 30$, $I = 6.0\ \text{A}$, $B = 1.0\ \text{T}$, and $\theta$ is the angle between the field and the coil's normal. Here $\theta = 60^\circ$.

Area of the circular coil:

Substitute:

$\tau = 30 \times 6.0 \times 0.0201 \times 1.0 \times \sin 60^\circ$ …

Why magnetic torque?

A current-carrying loop in a magnetic field experiences a torque that tries to rotate it so that its normal aligns with the field. This torque arises because the field exerts opposite forces on opposite sides of the loop, creating a couple. The magnitude of this torque depends on the number of turns $N$, the current $I$, the area $A$ of the loop, the field strength $B$, and the angle $\theta$ between the field and the normal to the coil.

The formula is:

To keep the coil from turning, we must apply an equal and opposite torque — a counter torque — of the same magnitude.

Step-by-step solution

1. Identify the given data

• Number of turns: $N = 30$
• Radius of coil: $r = 8.0\ \text{cm} = 0.08\ \text{m}$
• Current: $I = 6.0\ \text{A}$
• Magnetic field: $B = 1.0\ \text{T}$
• Angle between field and normal: $\theta = 60^\circ$

2. Compute the area of the coil

The coil is circular, so its area is:

(Keep a few extra digits for accuracy; we'll round at the end.)

3. Apply the torque formula

Substitute the values:

We know $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.8660$.

So:

4. Calculate step by step

First, $30 \times 6.0 = 180$.

Then $180 \times 0.020106 = 3.61908$.

Finally, $3.61908 \times 0.8660 = 3.134\ \text{N·m}$ (rounded to three significant figures).

5. Answer for part (a)

The counter torque required is $3.13\ \text{N·m}$.

Part (b): Does shape matter? …

Method: Torque on a Current-Carrying Coil in a Magnetic Field

This problem uses the magnetic torque formula for a planar coil. The torque arises because the magnetic field exerts forces on the current-carrying wires, tending to rotate the coil.

Steps for part (a)

1. Recall the torque formula The magnitude of torque on a planar coil is:

where:

• $N$ = number of turns
• $I$ = current in the coil
• $A$ = area of the coil
• $B$ = magnetic field magnitude
• $\theta$ = angle between the magnetic field and the normal to the coil plane

2. Identify the given values

   • $N = 30$
   • $r = 8.0\ \text{cm} = 0.08\ \text{m}$
   • $I = 6.0\ \text{A}$
   • $B = 1.0\ \text{T}$
   • $\theta = 60^\circ$ (angle between field and normal)

3. Calculate the area

4. Compute the torque

5. Counter torque required …

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong formula for torque

The error: Students often use $\tau = N I B A \sin\theta$ but confuse which angle $\theta$ represents. Some use $\theta = 60^\circ$ directly, while others mistakenly take $\theta = 30^\circ$ without reasoning.

Why it happens: The problem states: "field lines make an angle of $60^\circ$ with the normal". The torque formula uses the angle between $\vec{B}$ and the normal $\hat{n}$ — that is exactly $60^\circ$ here.

How to avoid: Always write the torque formula clearly:

where $\phi$ is the angle between $\vec{B}$ and the normal to the coil.

• If the problem says "angle with the normal" → that's your $\phi$ directly.
• If it says "angle with the plane of the coil" → subtract from $90^\circ$ to get $\phi$.

✓ Here: $\phi = 60^\circ$, so $\sin 60^\circ = \frac{\sqrt{3}}{2}$.

Mistake 2: Forgetting to convert radius to area correctly

The error: Using $r = 8.0\ \text{cm}$ directly in the formula without converting to metres, or using $A = \pi r$ instead of $A = \pi r^2$.

How to avoid:

• Always convert cm → m: $8.0\ \text{cm} = 0.08\ \text{m}$
• Area of a circle: $A = \pi r^2 = \pi (0.08)^2 = 0.0201\ \text{m}^2$ (approx)
• Keep the exact expression: $A = \pi \times 64 \times 10^{-4} = 64\pi \times 10^{-4}\ \text{m}^2$

Mistake 3: Forgetting the number of turns $N$

The error: Using $\tau = I B A \sin\theta$ instead of $\tau = N I B A \sin\theta$.

Why it happens: Students recall the formula for a single loop and forget that $N$ turns multiply the torque.

How to avoid: Write the full formula every time:

Here $N = 30$, so the torque is 30 times that of a single turn.

Mistake 4: Misinterpreting "counter torque"

The error: Students calculate the magnetic torque and stop there, forgetting that the question asks for the counter torque needed to prevent turning.

How to avoid:

• The magnetic torque tries to rotate the coil.
• To keep it stationary, you must apply an equal and opposite torque.
• So the magnitude of the required counter torque = magnitude of magnetic torque.

✓ Answer for (a): $\tau_{\text{counter}} = N I B A \sin 60^\circ$

Mistake 5: Thinking shape matters for part (b)

The error: Students think an irregular shape changes the torque calculation.

Why it happens: Intuition says "irregular" must be different — but physics says otherwise. …