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# A coil is placed in  a vertical plane and it 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 intially the axis of the coi is in the direction of the field . The coil rotates through an angle of 90 degree under the influence of  the magnetic field. ( a) What are the magnitudes of the torques on the coil in the initial and final positions? ( b) What is the angular speed acquired by the coil when it has rotated by 90 degree ? The moment of the coil is 0.1 kgm ?

Ram Chaudhary
13 Points
2 years ago
(a) Initally magnetic field and magnetic dipole moment are in at 0 degree angle
Hence Torque= M B sin(Theta)= 0
(b) Now angle between M and B is 90 degree hence Torque = M B sin(90)=M B
(c) ${}\frac{I\omega ^{2}}{2}=\int_{0}^{\prod /2} MBsin(\Theta )d\Theta$