A closely wound solenoid of 2000 turns and area of cross section `1.6xx10^-4m^2`, carrying a current of `4amp.` 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 are the force and torque on the solenoid if a uniform horizontal magnetic field of `7*5xx10^-2T` is set up at an angle of `30^@` with the axis of the solenoid?

A closely wound solenoid of 2000 turns and area of cross section `1.6x

1.	A closely wound solenoid of 2000 turns and area of cross section `1.6xx10^-4m^2`, carrying a current of `4amp.` 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 are the force and torque on the solenoid if a uniform horizontal magnetic field of `7*5xx10^-2T` is set up at an angle of `30^@` with the axis of the solenoid?
Answer» Number of turns on the solenoid, n=2000 Area of cross-section of the solenoid, A=`1.6xx10^(-4)m^(2)` Current in the solenoid, I=4A (a) The magnetic moment along the axis of the solenoid is calculated as: M=nAI `=2000xx1.6xx10^(-4)xx4` `=1.28Am^(2)` (b) Mangetic field `B=7.5xx10^(-2)T` angle between the magnetic field and the axis of the solenoid, `theta=30^(@)` Torque, r=`MB sin theta` `=1.28xx7.5xx10^(-7)sin30^(@)` `=4.8xx10^(-2)Nm` Since the magnetic field is uniform, the force on the solenoid is zero, the torque on the solenoid is `4.8xx10^(-2)Nm`.

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