Q.A closely wound solenoid of $2000$ turns and area of cross-section $1.6 \times 10^{-4}\ \text{m}^2$, carrying a current of $4.0\ \text{A}$, is suspended through its centre allowing it to turn in a horizontal plane.

Concept: Magnetic moment of a current-carrying solenoid; torque on a magnetic dipole in a uniform field.

(a) The magnetic moment of a solenoid is

$M = N I A$

where $N = 2000$, $I = 4.0\ \text{A}$, $A = 1.6 \times 10^{-4}\ \text{m}^2$.

$M = 2000 \times 4.0 \times 1.6 \times 10^{-4} = 1.28\ \text{A m}^2$.

(b) In a uniform magnetic field, the net force on a closed current loop is zero.

Torque is $\tau = M B \sin\theta$, with $\theta = 30^\circ$ between the solenoid axis and the field. …

This problem tests two core ideas from magnetism: first, that a current-carrying solenoid behaves like a bar magnet with a well-defined magnetic moment; second, how that magnetic moment interacts with an external uniform field. The key is to see the solenoid as a collection of current loops stacked together — each loop contributes its own magnetic moment, and they all add up.

The magnetic moment of a single turn is $I A$, where $I$ is the current and $A$ the area. For $N$ identical turns, the total moment is simply $N$ times that. That’s the first part.

For the second part, a uniform magnetic field exerts no net force on a magnetic dipole (the solenoid), because the field is the same everywhere — the forces on opposite sides cancel. But it does exert a torque that tries to align the solenoid’s axis with the field. The torque magnitude depends on the moment, the field strength, and the sine of the angle between them.

Let’s work through it.

1. Magnetic moment of the solenoid The formula for the magnetic moment of a planar current loop is $M = I A$ for one turn. For $N$ turns closely wound, the moments add directly because each turn carries the same current and has the same area.

Substitute: $N = 2000$, $I = 4.0\ \text{A}$, $A = 1.6 \times 10^{-4}\ \text{m}^2$.

2. Force on the solenoid in a uniform field A uniform magnetic field means $\vec{B}$ has the same magnitude and direction at every point. For a magnetic dipole (like our solenoid), the net force in a uniform field is always zero. Why? Because the field exerts equal and opposite forces on the north and south poles of the equivalent magnet — they cancel exactly. $\vec{F}_{\text{net}} = 0$ …

Method: Magnetic Moment and Torque on a Current-Carrying Solenoid

This problem uses the magnetic dipole model for a solenoid — treating it as a magnetic dipole with a well-defined magnetic moment.

(a) Magnetic Moment of the Solenoid

Step 1: Recall the formula

For a closely wound solenoid, the magnetic moment is:

where:

• $N$ = number of turns
• $I$ = current
• $A$ = cross-sectional area
• $\hat{n}$ = unit vector along the solenoid axis

Step 2: Substitute values

Step 3: Result

Magnetic moment = $\mathbf{1.28 \ A\,m^2}$, directed along the solenoid axis.

(b) Force and Torque in a Uniform Magnetic Field

Step 1: Force on the solenoid

In a uniform magnetic field, the net force on a magnetic dipole is zero:

Step 2: Torque on the solenoid

Torque on a magnetic dipole in a uniform field:

Magnitude:

where $\theta$ is the angle between $\vec{m}$ (solenoid axis) and $\vec{B}$.

Step 3: Substitute values

• $m = 1.28 \ \text{A m}^2$
• $B = 7.5 \times 10^{-2} \ \text{T}$
• $\theta = 30^\circ$

$\tau = 1.28 \times 0.075 \times 0.5$ …

🔍 Mistake 1: Wrong formula for magnetic moment of a solenoid

What students do wrong:

They use $M = NIA$ but forget that for a solenoid, the magnetic moment is:

where $N$ is the total number of turns, $I$ is current, $A$ is cross-sectional area, and $\hat{n}$ is the unit vector along the solenoid axis.

How to avoid:

Always write the definition:

Magnetic moment = (number of turns) × (current) × (area per turn)

Not current × area alone.

For part (a):

✓ Correct answer: $\mathbf{1.28 \ A\ m^2}$

🔍 Mistake 2: Confusing force and torque on a magnetic dipole in a uniform field

What students do wrong:

They think a net force acts on the solenoid in a uniform field.

How to avoid:

Remember:

• In a uniform magnetic field, net force on a magnetic dipole (like a solenoid) is zero.
• Only torque exists.

So for part (b):

✓ Correct answer: $\mathbf{0 \ N}$

🔍 Mistake 3: Using wrong angle in torque formula

What students do wrong:

They use $\tau = mB \sin\theta$ but take $\theta = 30^\circ$ directly from the problem statement — which is correct here, but they often forget that $\theta$ is the angle between $\mathbf{m}$ and $\mathbf{B}$.

How to avoid:

Always check:

$\theta$ = angle between magnetic moment vector and magnetic field vector.

Here, the problem says: field is at $30^\circ$ with the axis of the solenoid — and the axis is the direction of $\mathbf{m}$. So $\theta = 30^\circ$ is correct.

Torque magnitude:

✓ Correct answer: $\mathbf{0.048 \ N m}$

🔍 Mistake 4: Forgetting the direction of torque

What students do wrong:

They give only magnitude, but the problem asks for “force and torque” — direction matters. …