Question

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6 A is suspended vertically in a uniform horizontal magnetic field of 1.0 T. The field lines make an angle of $30^\circ$ with the plane of the coil. Calculate the magnitude of the external torque that must be applied to prevent the coil from turning. What would happen if the circular coil is replaced by a planar coil of irregular shape that encloses the same area, keeping other parameters unchanged ?

Hint:

The most common mistake here is using the angle given directly ($30^\circ$). Always remember that the formula uses $\sin(\theta)$ where $\theta$ is the angle with the NORMAL to the plane. If the angle with the plane is $\alpha$, use $\theta = 90^\circ - \alpha$.

Answer 1

Concept:
• A current-carrying coil placed in a magnetic field experiences a magnetic torque.
• The magnitude of this torque is given by $\tau = N I A B \sin(\theta)$.
• Here, $N$ is the number of turns, $I$ is the current, $A$ is the cross-sectional area, $B$ is the magnetic field strength, and crucially, $\theta$ is the angle between the magnetic field vector and the area vector (which is the normal to the plane of the coil).
• To prevent the coil from turning, an external torque of exactly equal magnitude must be applied in the opposite direction.

Step 1: Extract the given parameters
Number of turns, $N = 30$.
Radius of the coil, $r = 8.0 \text{ cm} = 0.08 \text{ m}$.
Current, $I = 6 \text{ A}$.
Magnetic field, $B = 1.0 \text{ T}$.
The angle the magnetic field makes with the plane of the coil is given as $30^\circ$.
Therefore, the angle $\theta$ between the normal to the coil (area vector) and the magnetic field is $\theta = 90^\circ - 30^\circ = 60^\circ$.

Step 2: Calculate the area of the circular coil
The area $A$ of a circle is $\pi r^2$.
\[ A = \pi \times (0.08 \text{ m})^2 \]
\[ A = \pi \times 0.0064 \text{ m}^2 \]

Step 3: Calculate the magnitude of the torque
Apply the magnetic torque formula:
\[ \tau = N I A B \sin(\theta) \]
\[ \tau = 30 \times 6 \times (\pi \times 0.0064) \times 1.0 \times \sin(60^\circ) \]
\[ \tau = 180 \times 0.0064\pi \times \frac{\sqrt{3}}{2} \]
\[ \tau = 1.152\pi \times 0.866 \]
\[ \tau = 3.133 \text{ N m} \]
The magnitude of the external torque required to balance this is exactly $3.13 \text{ N m}$.

Step 4: Analyze the replacement with an irregular coil
The formula for magnetic torque ($\tau = N I A B \sin\theta$) solely depends on the total enclosed area $A$ of the coil, not on its geometric shape.
The problem explicitly states that the new irregular planar coil encloses the same area and all other parameters ($N, I, B, \theta$) remain unchanged.
Therefore, the magnetic moment ($M = N I A$) remains identical.

Step 5: Conclusion
The required external torque is $3.13 \text{ N m}$. If replaced by an irregular coil of the same area, the torque experienced by the coil would remain exactly the same.