Magnetism E XAMPLE . equilibrium is a stable one. Any small rotation of the coil produces a torque which brings it back to its original position. When they are antiparallel, the equilibrium is unstable as any rotation produces a torque which increases with the amount of rotation.
The presence of this torque is also the reason why a small magnet or any magnetic dipole aligns itself with the external magnetic field. If the loop has N closely wound turns, the expression for torque, Eq. ( . ), still holds, with m = N I A ( .
) Example . A turn closely wound circular coil of radius cm carries a current of . 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 90º 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 90º? The moment of inertia of the coil is . kg m .
Solution (a) From Eq. ( . ) NI B µ Here, N = ; I = . A, and R = .
m. Hence, . B π × (using π × . = ) = × – T The direction is given by the right-hand thumb rule.
(b) The magnetic moment is given by Eq. ( . ), m = N I A = N I π r = × . × .
× – = A m The direction is once again given by the right hand thumb rule. (c) τ = B [from Eq. ( . )] sin m B θ Initially, θ = .
Thus, initial torque τ i = . Finally, θ = π / (or 90º). Thus, final torque τ f = m B = × = N m. (d) From Newton’s second law, d sin m B θ where is the moment of inertia of the coil.
From chain rule, θ ω ω θ θ Using this, sin m B θ θ E XAMPLE . E XAMPLE . Integrating from θ = to θ = π / , / sin f m B θ θ