Question

A circular coil of \(100\) turns and radius \[ \left(\frac{10}{\pi}\right)\text{ cm} \] carrying current \(5.0\) A is suspended vertically in a uniform horizontal magnetic field of \(2.0\) T. The field makes an angle \(30^\circ\) with the normal to the coil. Calculate:
(I) the magnetic dipole moment of the coil, and
(II) the magnitude of the counter torque required to prevent the coil from turning.

Answer 1

Numerical Calculation

Step 1: Calculate area of the circular coil. Radius: \[ r=\frac{10}{\pi}\text{ cm} = \frac{0.10}{\pi}\text{ m}. \] Area of one turn: \[ A=\pi r^2. \] \[ A = \pi\left(\frac{0.10}{\pi}\right)^2. \] \[ A = \frac{0.01}{\pi}. \] \[ A = 3.183\times10^{-3}\,\text m^2. \]

Step 2: Calculate magnetic dipole moment. \[ m=NIA. \] Given \[ N=100, \qquad I=5\,A. \] Hence \[ m = 100\times5\times3.183\times10^{-3}. \] \[ m = 1.59\,\text{A m}^2. \] Therefore, \[ \boxed{ m=1.59\,\text{A m}^2 } \]

Step 3: Calculate torque on the coil. \[ \tau=mB\sin\theta. \] Given \[ B=2.0\,T, \qquad \theta=30^\circ. \] Therefore, \[ \tau = (1.59)(2)(\sin30^\circ). \] \[ \tau = 1.59\times2\times\frac12. \] \[ \tau=1.59\,\text{N m}. \] Hence the counter torque required is \[ \boxed{ 1.59\,\text{N m} } \]