A short bar magnet placed with its axis at 45^ with an external field of 400 10⁻⁴ ~T experiences a torque of 0.024 Nm . If a solenoid of cross-sectional area 10⁻⁴ ~m ^2 and 500 turns replaces the short bar magnet such that they have the same magnetic moment, then the current flowing through the solenoid is:

Step-by-step solution

The torque experienced by a magnetic dipole in an external magnetic field B is given by = mB , where m is the magnetic moment and is the angle between the magnetic moment and the magnetic field. Given = 0.024 Nm , B = 400 10⁻⁴ T = 0.04 T , and = 45^ . Substituting the values: 0.024 = m 0.04 (45^ ) . 0.024 = m 0.04 1 2 . m = 0.024 2 0.04 = 0.6 2 Am ^2 . The magnetic moment of a solenoid is given by m = NIA , where N is the number of turns, I is the current, and A is the cross-sectional area. Given N = 500 and A = 10