A bar magnet has a magnetic moment 200 Am2. The magnet is suspended in a magnetic field of 0.30 NA-1m-1. The torque required to rotate the magnet from its equilibrium position at an angle of 30˚ will be:

Correct answer: A. 30Nm

  • A. 30Nm
  • B. 3073 Nm
  • C. 60Nm
  • D. 604Nm

Explanation

τ = MBsinθTorque = 200×0.30×sin30 = 30 Nm.

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