Q.A square coil of side $10\ \text{cm}$ consists of $20$ turns and carries a current of $12\ \text{A}$. The coil is suspended vertically and the normal to the plane of the coil makes an angle of $30^\circ$ with the direction of a uniform horizontal magnetic field of magnitude $0.80\ \text{T}$. What is the magnitude of torque experienced by the coil?

Concept: Magnetic Force Balance — torque on a current-carrying coil in a uniform field.

Step 1: Formula for torque

For a planar coil of $N$ turns, area $A$, carrying current $I$, in a uniform field $B$, the magnitude of torque is

where $\theta$ is the angle between the normal to the coil and the magnetic field.

Step 2: Insert values

Side $= 10\ \text{cm} = 0.10\ \text{m}$, so $A = (0.10)^2 = 0.01\ \text{m}^2$. …

Why Magnetic Torque?

When a current loop sits in a magnetic field, each side of the loop experiences a magnetic force. If the loop is oriented so that its plane is not parallel to the field, these forces produce a torque that tries to rotate the loop. For a coil with multiple turns, the torque simply multiplies by the number of turns.

The key formula is:

Notice: the problem gives the angle between the normal and the field directly — that’s exactly the $\theta$ we need. No conversion required.

Step-by-step solution

1. Identify the given quantities

   Side length of square coil: $l = 10\ \text{cm} = 0.10\ \text{m}$

   Number of turns: $N = 20$

   Current: $I = 12\ \text{A}$

   Magnetic field: $B = 0.80\ \text{T}$

   Angle between normal and field: $\theta = 30^\circ$

2. Compute the area of the coil

   Since it’s a square:

3. Apply the torque formula

Substitute the values:

4. Simplify step by step

   $\sin 30^\circ = 0.5$

   First multiply the constants: $20 \times 12 = 240$

   Then $240 \times 0.01 = 2.4$

   Then $2.4 \times 0.80 = 1.92$

   Finally $1.92 \times 0.5 = 0.96$

   So $\tau = 0.96\ \text{N·m}$. …

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

This problem uses the torque on a current loop formula, derived from the magnetic force balance on each side of the coil.

Steps

Step 1: Identify the formula

The torque $\tau$ on a planar coil in a uniform magnetic field is given by:

where:

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

Step 2: Extract given values

• Side of square = $10\ \text{cm} = 0.10\ \text{m}$
• Area $A = (0.10)^2 = 0.01\ \text{m}^2$
• $N = 20$ turns
• $I = 12\ \text{A}$
• $B = 0.80\ \text{T}$
• $\theta = 30^\circ$

Step 3: Substitute into the formula

Step 4: Simplify

• $\sin 30^\circ = 0.5$
• $20 \times 12 = 240$ …

Here are the common mistakes students make on this Magnetic Torque problem and how to avoid each.

1. Using the wrong formula for area

Mistake:

Students often use the side length directly in the torque formula without first calculating the area of the coil.

Why it’s wrong:

Torque depends on the area of the coil ($A$), not just its side length. For a square, $A = (\text{side})^2$.

How to avoid:

Always write:

Convert cm to m before squaring.

2. Forgetting to multiply by the number of turns ($N$)

Mistake:

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

Why it’s wrong:

Each turn experiences torque. The total torque is the sum over all turns.

How to avoid:

Memorise the correct formula:

Check the problem statement for $N$ — here $N = 20$.

3. Using the wrong angle in $\sin\theta$

Mistake:

Taking $\theta = 30^\circ$ directly as the angle between $\vec{B}$ and the plane of the coil.

Why it’s wrong:

The formula uses the angle between the magnetic field and the normal to the coil’s plane. The problem gives exactly that: $30^\circ$ between normal and $\vec{B}$.

How to avoid:

• Read carefully: “normal to the plane makes an angle of $30^\circ$ with the direction of the field”
• That is the $\theta$ in $\tau = N I A B \sin\theta$
• No conversion needed — use $\sin 30^\circ = 0.5$ directly.

4. Unit conversion errors

Mistake:

Using side = 10 cm as 10 m, or forgetting to convert cm² to m².

Why it’s wrong:

Torque in SI units requires metres, not centimetres.

How to avoid:

Convert all lengths to metres before any calculation:

Then area:

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