Question

A coil of 60 turns and area \( 1.5 \times 10^{-3} \, \text{m}^2 \) carrying a current of 2 A lies in a vertical plane. It experiences a torque of 0.12 Nm when placed in a uniform horizontal magnetic field. The torque acting on the coil changes to 0.05 Nm after the coil is rotated about its diameter by 90°. Find the magnitude of the magnetic field.

Hint:

The torque on a current-carrying coil depends on the number of turns, current, area, magnetic field strength, and the angle between the field and the normal to the coil.

Answer 1

Concept: Magnetic moment of the coil is $$\mu = nIA.$$ Torque on the coil in a magnetic field $B$ is $$\tau=\mu B\sin\phi$$ where $\phi$ is the angle between $\mu$ (normal to the coil) and $\mathbf{B}$. 
When the coil is rotated about a suitable diameter by $90^\circ$, the angles change so that the two measured torques satisfy $$\tau_1=\mu B\sin\phi,\qquad \tau_2=\mu B\cos\phi.$$ Squaring and adding these two equations gives $$\tau_1^2+\tau_2^2=(\mu B)^2.$$ Hence $$\mu B=\sqrt{\tau_1^2+\tau_2^2}.$$ Therefore $$B=\dfrac{\sqrt{\tau_1^2+\tau_2^2}}{\mu}=\dfrac{\sqrt{\tau_1^2+\tau_2^2}}{nIA}.$$ 

Calculation:
$$\sqrt{\tau_1^2+\tau_2^2}=\sqrt{(0.12)^2+(0.05)^2}=\sqrt{0.0144+0.0025}=\sqrt{0.0169}=0.13\ \mathrm{N\,m}.$$
$$\mu = nIA = 60\times 2\times 1.5\times10^{-3}=0.18\ \mathrm{A\,m^2}.$$ Thus $$B=\dfrac{0.13}{0.18}=\dfrac{13}{18}\ \mathrm{T}\approx 0.7222\ \mathrm{T}.$$ 

Answer: $\displaystyle B=\frac{13}{18}\ \mathrm{T}\approx 0.722\ \mathrm{T}.$

Answer 2

The torque \( \tau \) on a current-carrying coil in a magnetic field is given by: \[ \tau = n I A B \sin \theta \]

where:
\( n \) is the number of turns of the coil,
- \( I \) is the current,
- \( A \) is the area of the coil,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the magnetic field and the normal to the coil.
Initially, when the coil is in the vertical plane (\( \theta = 90^\circ \)), the torque is: \[ \tau_1 = n I A B \sin 90^\circ = n I A B \] Substituting the known values: \[ 0.12 = 60 \times 2 \times 1.5 \times 10^{-3} \times B \] Solving for \( B \): \[ B = \frac{0.12}{60 \times 2 \times 1.5 \times 10^{-3}} = 0.67 \, \text{T} \] Thus, the magnitude of the magnetic field is \( 0.67 \, \text{T} \).