InterviewSolution
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InterviewSolution
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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 unifrom magnetic field of 2 T 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 acqired by the coil when it has rotated by 90^(@) ? The moment of inertia of the coil is 0.1 kg m^2 |
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Answer» Solution :`B = (mu_0 NI)/(2R)` Here, N =100, I = 3.2 A, and R = 0.1 m . Hence, `B = (4 PI 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 `m = NI A = NI pi r^2 = 100 xx 3.2 xx 3.14 xx 10^(-2) = 10 Am^2` The direction is given by the right hand thumb rule . c. `tau = | m xx B| = mB sin theta` Initially, `theta = theta` Thus, inital torque `tau_i =0` Finally , `theta = (pi)/(2) ( or 90^@)` Thus , FINAL torque ` tau_t = mB =10 xx 2 = 20 Nm ` d. Prom Newton.s second law, `I (d omega)/(dt) = m B sin theta` where I is the moment of inertia of the coil. From chain rule, `(domega)/(dt)=(domega)/(d theta)(d theta)/(dt)=(d omega)/(d theta) omega` Using this , `I omega d omega = m B sin theta d theta ` Intergrating from `theta = 0 " to " theta = pi/2 ,1 int_(0)^(OMEGAR) omegadomega = mBint_(0)^(pi//2)sin theta d theta` `I (omega_r^2)/(2) =- mB COS theta "|"_(0)^(pi //2) = m B` `omega_(f) = ((2mB)/(I))^(1/2) = ((2 xx20)/(10^(-1))) ^(1/2) = 20 s^(-1)` |
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