Q.If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25\ \text{T}$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^\circ$ with the direction of applied field?

Concept: Magnetic Poles — a current-carrying solenoid behaves like a bar magnet with a magnetic moment $\vec{m}$.

Step 1: From Exercise 5.5, the magnetic moment of the solenoid is

$m = 0.6\ \text{J/T}$.

Step 2: Torque on a magnetic dipole in a uniform field is

$\tau = mB \sin\theta$, where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$.

Step 3: Here $B = 0.25\ \text{T}$, $\theta = 30^\circ$, so …

Step-by-Step Solution

Magnetic moment of the solenoid (from the Exercise 5.5 data $N=800$, $I=3.0\ \text{A}$, $A=2.5\times10^{-4}\ \text{m}^2$):

Torque on a magnetic moment in a uniform field:

$\tau=mB\sin\theta,$ …

Method: Torque on a Magnetic Dipole in a Uniform Magnetic Field

This problem uses the magnetic dipole torque formula — the solenoid behaves like a bar magnet with a magnetic moment.

Step-by-step solution

Step 1: Recall the torque formula

For a magnetic dipole (or solenoid) in a uniform magnetic field:

Magnitude:

where:

• $M$ = magnetic moment of the solenoid
• $B$ = applied magnetic field strength
• $\theta$ = angle between the solenoid axis and the field direction

Step 2: Identify given values

From Exercise 5.5 (assumed context), the magnetic moment of the solenoid is:

Given in this problem:

• $B = 0.25\ \text{T}$
• $\theta = 30^\circ$

Step 3: Apply the formula

Since $\sin 30^\circ = \frac{1}{2}$: …

Here are the common mistakes students make on this magnetic torque problem, along with how to avoid each.

1. Using the Wrong Formula for Torque

Mistake:

Students often apply $\tau = mB$ (the formula for force on a current-carrying wire) or $\tau = NIAB$ without the $\sin\theta$ factor.

Why it’s wrong:

Torque on a magnetic dipole (like a solenoid) in a uniform field is given by:

where $m = NIA$ is the magnetic moment of the solenoid.

How to avoid:

Always write the full vector form first: $\vec{\tau} = \vec{m} \times \vec{B}$. Then extract magnitude as $\tau = mB \sin\theta$. Never drop the $\sin\theta$ term.

2. Confusing the Angle $\theta$

Mistake:

Using $\theta = 30^\circ$ directly in $\sin\theta$ when the problem says the axis makes $30^\circ$ with the field.

Why it’s correct here:

The angle between the magnetic moment vector (along the solenoid axis) and the field is exactly $30^\circ$. So $\theta = 30^\circ$ is correct.

How to avoid:

Draw a diagram. Mark the solenoid axis, the field direction, and the angle between them. That angle is $\theta$ in $\tau = mB\sin\theta$.

3. Forgetting to Calculate $m = NIA$ First

Mistake:

Plugging numbers directly into $\tau = NIAB\sin\theta$ without computing $m$ — or using wrong values for $N$, $I$, or $A$.

Why it’s wrong:

The torque depends on the product $NIA$, not just $I$ or $A$ alone.

How to avoid:

List given data:

• $N$ = number of turns
• $I$ = current
• $A$ = cross-sectional area ($\pi r^2$ for circular)

Then compute $m = NIA$ separately before plugging into torque formula.

4. Unit Errors (Area, Field, Torque)

Mistake:

Using area in $\text{cm}^2$ without converting to $\text{m}^2$, or mixing Tesla with Gauss.

How to avoid:

Always convert to SI units:

• $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$
• $1\ \text{G} = 10^{-4}\ \text{T}$ …