Q.A $100$ turn closely wound circular coil of radius $10\ \text{cm}$ carries a current of $3.2\ \text{A}$.

Concept: Magnetic Force Balance — the torque on a current loop in a uniform field is $\vec{\tau} = \vec{m} \times \vec{B}$, and work done by the field equals the change in rotational kinetic energy.

(a) Field at centre of a circular coil:

$B = \frac{\mu_0 N I}{2R}$

$= \frac{4\pi \times 10^{-7} \times 100 \times 3.2}{2 \times 0.10} = 2.01 \times 10^{-3}\ \text{T}$.

(b) Magnetic moment: $m = N I A = N I (\pi R^2)$

$= 100 \times 3.2 \times \pi \times (0.10)^2 = 10.05\ \text{A m}^2$.

(c) Torque magnitude: $\tau = mB \sin\theta$

Initial ($\theta = 0$): $\tau_i = 0$

Final ($\theta = 90^\circ$): $\tau_f = 10.05 \times 2 = 20.1\ \text{N m}$.

(d) Work done by field = $mB(\cos 0 - \cos 90^\circ) = mB = 20.1\ \text{J}$

This equals $\frac{1}{2} I \omega^2$:

$\omega = \sqrt{\frac{2 \times 20.1}{0.1}} = \sqrt{402} \approx 20.05\ \text{rad/s}$.

Concept and Intuition

This is a beautiful blend of three ideas: the magnetic field produced by a current-carrying coil, the magnetic moment that governs how the coil interacts with an external field, and the resulting torque that makes it rotate. The key is that the torque depends on the angle between the coil’s magnetic moment and the external field — it’s maximum when they’re perpendicular, zero when aligned. When the coil is free to rotate, the torque does work, converting magnetic potential energy into rotational kinetic energy. That energy conservation gives us the final angular speed.

Step-by-step solution

1. Field at the centre of the coil

For a circular coil of $N$ turns, radius $R$, carrying current $I$, the magnetic field at the centre is:

Here $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$, $N = 100$, $I = 3.2\ \text{A}$, $R = 0.10\ \text{m}$.

Simplify: $320 / 0.20 = 1600$, so

Numerically, $\pi \approx 3.1416$, so

2. Magnetic moment of the coil

Magnetic moment $m$ for a planar coil is:

where $A = \pi R^2$ is the area.

So

Numerically, $m \approx 10.05\ \text{A m}^2$.

3. Torque on the coil in initial and final positions

Torque on a magnetic dipole in a uniform field is:

where $\theta$ is the angle between $m$ and $B$.

Initial position: The axis of the coil (direction of $m$) is aligned with the external field $B$. So $\theta = 0^\circ$, $\sin 0 = 0$, hence

Final position: The coil rotates by $90^\circ$, so $m$ becomes perpendicular to $B$. Then $\theta = 90^\circ$, $\sin 90^\circ = 1$, so

Numerically, $\tau_{\text{final}} \approx 20.1\ \text{N m}$.

4. Angular speed after rotating $90^\circ$

The coil is free to rotate, and the magnetic field does work on it. The change in magnetic potential energy equals the gain in rotational kinetic energy.

Magnetic potential energy for a dipole is:

Initially, $\theta = 0^\circ$, so $U_i = - m B$.

Finally, $\theta = 90^\circ$, so $U_f = 0$.

The loss in potential energy is:

This becomes kinetic energy:

Given $I = 0.1\ \text{kg m}^2$, $m = 3.2\pi\ \text{A m}^2$, $B = 2\ \text{T}$:

Numerically, $\sqrt{128\pi} \approx \sqrt{402.12} \approx 20.05\ \text{rad/s}$.

So $\omega \approx 20.0\ \text{rad/s}$.

Method: Magnetic Force Balance & Torque on a Current Loop

This problem uses the torque on a magnetic dipole in a uniform field method. The key idea: a current-carrying coil behaves like a magnetic dipole with moment $\vec{m}$, experiencing torque $\vec{\tau} = \vec{m} \times \vec{B}$.

Step-by-step solution

(a) Magnetic field at centre of coil

For a circular coil of $N$ turns, radius $R$, carrying current $I$:

Given: $N = 100$, $R = 0.10\ \text{m}$, $I = 3.2\ \text{A}$, $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$

$B = 2.01 \times 10^{-3}\ \text{T}$ (or $2.01\ \text{mT}$)

(b) Magnetic moment of the coil

Magnetic moment: $\vec{m} = N I \vec{A}$, where $A = \pi R^2$ is area vector (direction along coil axis)

$m = 10.05\ \text{A m}^2$

(c) Torque in initial and final positions

Torque magnitude: $\tau = mB \sin\theta$, where $\theta$ is angle between $\vec{m}$ and $\vec{B}$

• Initial position: axis of coil is along $\vec{B}$ → $\theta = 0^\circ$ → $\sin 0^\circ = 0$

$\tau_{\text{initial}} = 0$

• Final position: coil rotated by $90^\circ$ → $\theta = 90^\circ$ → $\sin 90^\circ = 1$

$\tau_{\text{final}} = 20.1\ \text{N m}$

(d) Angular speed after rotating $90^\circ$

Use work-energy theorem: work done by torque = change in rotational kinetic energy

Work done by torque from $\theta = 0^\circ$ to $\theta = 90^\circ$:

This work equals final kinetic energy: $\frac{1}{2} I \omega^2$

Given $I = 0.1\ \text{kg m}^2$:

$\omega = 20.05\ \text{rad/s}$

Key Concept Summary

Here’s a breakdown of the common mistakes students make on this magnetic force balance problem and how to avoid each.

1. Using the wrong formula for the magnetic field at the centre

Mistake:

Students often use the formula for the field at the centre of a straight solenoid or a long wire instead of the correct one for a circular coil.

Correct approach:

For a circular coil of $N$ turns, radius $r$, carrying current $I$, the field at the centre is:

How to avoid:

• Memorise the formula with the factor $2r$ in the denominator (not $r$ or $4\pi r$).
• Check units: $B$ comes out in tesla (T) when $r$ is in metres and $I$ in amperes.

2. Forgetting to convert cm to m

Mistake:

Using $r = 10$ directly instead of $r = 0.10\ \text{m}$.

How to avoid:

• Always write the conversion step: $10\ \text{cm} = 0.10\ \text{m}$.
• Double-check every length in the problem before plugging into a formula.

3. Confusing magnetic moment with magnetic field

Mistake:

Writing $\vec{m} = N I \vec{B}$ or using the field value to compute moment.

Correct formula:

Magnetic moment of a planar coil:

where $A = \pi r^2$ is the area.

How to avoid:

• Remember: moment depends only on current, turns, and area — not on the external field.
• Write the definition: $\vec{m} = N I \vec{A}$ (direction given by right-hand rule).

4. Torque sign and magnitude errors

Mistake:

Writing $\tau = mB$ for all angles, or forgetting that torque depends on $\sin\theta$.

Correct:

Torque on a magnetic dipole in a uniform field:

where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$.

• Initial position: axis of coil (and hence $\vec{m}$) is parallel to $\vec{B}$ → $\theta = 0$ → $\tau = 0$
• Final position: axis is perpendicular to $\vec{B}$ → $\theta = 90^\circ$ → $\tau = mB$

How to avoid:

• Draw a diagram showing $\vec{m}$ and $\vec{B}$ at each position.
• Use $\sin\theta$, not $\cos\theta$.

5. Using work-energy theorem incorrectly for angular speed

Mistake:

Assuming $\tau$ is constant and using $\tau \theta = \frac{1}{2} I \omega^2$ with $\theta$ in degrees.

Correct approach:

Torque varies with angle: $\tau(\theta) = mB \sin\theta$.

Work done by magnetic torque from $\theta_1$ to $\theta_2$:

Then use work-energy theorem:

So:

How to avoid:

• Always integrate $\tau \, d\theta$ when torque is not constant.
• Convert angles to radians only if using $\tau = mB \sin\theta$ with $\theta$ in radians — but here the integration is straightforward in degrees because $\sin\theta$ is dimensionless.

6. Forgetting the moment of inertia is given

Mistake:

Trying to compute $I$ from mass and geometry, or leaving $\omega$ in terms of $I$.

How to avoid:

• Read the problem carefully: $I = 0.1\ \text{kg m}^2$ is given. Use it directly.

Quick summary table

Final tip:

Always sketch the coil, its axis, and the external field direction. This makes the angle $\theta$ and the torque direction obvious — and prevents most of these errors.