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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 :(a) MAGNETIC moment of current carrying solenoid, `m_(s) = NIA` `therefore m_(s) = (2000) (4) (1.6 XX 10^(-4) )` `=1.28 Am^(2)` (b) (i) Resultant force exerted on a magnet in a UNIFORM exerted magnetic field is, `overset(to)(F) = -q_(m) overset(to)(B) + q_(m) overset(to) (B) = overset(to) (0)` (Where, `q_(m)=` pole strength of either pole of a magnet) (II) Torque exerted on above solenoid, `tau = m_(s) B sin theta` `= (1.28) (7.5 xx 10^(-2) ) (sin 30^(@) )` `= (1.28 ) (0.075 ) (0.5)` `= 0.048` Nm |
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