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A closely wound solenoid of 2000 turns and area of cross-section 1.6 xx 10^(-4)m^2 carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid ? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 xx 10^(-2) T is set up at an angle of 30^@ with the axis of the solenoid ? |
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Answer» SOLUTION :Here N = 2000 , A `=1.6 xx 10^(-4) m^2 and I = 0.4 A` (a) magnetic moment of solenoid m = NIA `= 2000xx4.0xx1.6xx10^(-4) = 1.28 m^2 ` (b) When B =` 7.5xx10^(-2) T and THETA = 30^@` Net force on the solenoid F= 0 [Since the magnetic field B is uniform one] Net torque on the solenoid `tau= m B SIN theta = 1.28 xx 7.5xx10^(-2) xx sin 30^@ = 4.8 xx 10^(-2) N m`. The Torque tends to align the solenoid AXIS along the direction of field `vecB`. |
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