MagnetismMedium
Question
A closely wound solenoid 200 turns area of cross section 1.5 × 10-4m2 carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turns in a horizontal plane in a uniform magnetic field 5 × 10-2 tesla making an angle 30o with the axis of the solenoid. The torque on the solenoid will be
Options
A.3 × 10-3Nm
B.1.5 × 10-3Nm
C.1.5 × 10-2Nm
D.3 × 10-2Nm
Solution
Magnetic moment of the loop
M = NIA = 200 × 2 × 1.5 × 10-4 = 0.6J/T
Torque
= MB sin 30o
= 0.6 × 5 × 10-2 ×
= 1.5 × 10-2 Nm
M = NIA = 200 × 2 × 1.5 × 10-4 = 0.6J/T
Torque
= MB sin 30o = 0.6 × 5 × 10-2 ×
= 1.5 × 10-2 NmCreate a free account to view solution
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