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: The magnetic moment of a coil is given by \(\mu = nIA\). The torque acting on the coil in a magnetic field \(B\) is \(\tau=\mu B\sin\phi\), where \(\phi\) is the angle between the magnetic moment vector \(\mu\) (normal to the coil) and the magnetic field vector \(\mathbf{B}\). When the coil is rotated by \(90^\circ\) about an axis, the measured torques are \(\tau_1=\mu B\sin\phi\) and \(\tau_2=\mu B\cos\phi\). Squaring and adding these equations yields \(\tau_1^2+\tau_2^2=(\mu B)^2\). Thus, \(\mu B=\sqrt{\tau_1^2+\tau_2^2}\). Consequently, \(B=\dfrac{\sqrt{\tau_1^2+\tau_2^2}}{\mu}=\dfrac{\sqrt{\tau_1^2+\tau_2^2}}{nIA}\).

Calculation:
The value of \(\sqrt{\tau_1^2+\tau_2^2}\) is calculated as \(\sqrt{(0.12)^2+(0.05)^2}=\sqrt{0.0144+0.0025}=\sqrt{0.0169}=0.13\ \mathrm{N\,m}\).
The magnetic moment \(\mu\) is \(nIA = 60\times 2\times 1.5\times10^{-3}=0.18\ \mathrm{A\,m^2}\).
Therefore, \(B=\dfrac{0.13}{0.18}=\dfrac{13}{18}\ \mathrm{T}\approx 0.7222\ \mathrm{T}\). 

Answer: The magnetic field strength is \(B=\dfrac{13}{18}\ \mathrm{T}\approx 0.722\ \mathrm{T}\).