A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. (a) What is the field at the centre of the coil? (b) What is the magnetic moment of this coil? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of `90^@` under the influence of the magnetic field. (c) What are the magnitudes of the torques on the coil in the initial and final position? (d) What is the angular speed acquired by the coil when it has rotated by `90^@`? The moment of inertia of the coil is` 0.1 kg m^2`.

Explanation

(a) From eq.
`B = (mu_0 NI)/(2 R)`
Here , N = 100 , I = 3.2 A, and R = 0.1 m . Hence
`B = (4pi xx 10^(-7) xx 10^2 xx 3.2)/(2 xx 10^(-1)) = (4 xx 10^(-5) xx 10)/(2 xx 10^(-1)) ` ( using `pi xx 3.2 = 10` )
The direction is given by the right-hand thumb rule.
(b) The magnetic moment is given by Eq.
`m = NIA = NI pi r^2 = 100 xx 3.2 xx 3.14 xx 10^(-2) = 10 A m^2`
The direction is once again given by the right-hand thumb rule
(c ) ` tau = | m xx B| `
` = m B sin theta `
Initially, ` theta = 0` . Thus , initial torque `tau_1 = 0` . finally , `theta = pi//2 ( or 90^@)`
Thus final torque `tau_f = mB = 10 xx 2 = 20 Nm`
(d) From newton.s second law
`(d omega)/(dt) = mB sin theta `
Where `omega` is the moment of inertia of the coil . from chain rule ,
`(d omega)/(dt) = (d omega)/(dtheta) ( d theta)/(dt) = (d omega)/(d theta) omega`
using this ,
`omega d omega =m B sin theta d theta `
Integrating from ` theta = 0 ` to ` theta =pi//2`
`g int_(0)^(omega_f) omega d omega = m B int_(0)^(pi//2) sin theta d theta`
` g (omega_f^2)/(2 ) = - mB cos theta ]_0^(pi//2) = mB`
`omega_f = ( (2m B)/( g) )^(1//2) = ((2 xx 20)/(10^(-1)))^(1//2) = 20 s^(-1)`