A short bar magnet placed with its axis at 30º with a uniform externalmagnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10-2J. What is the magnitude of magnetic moment of the magnet?
Magnetic field strength, B= 0.25 T
Torque on the bar magnet, T= 4.5 ×10-2J
Angle between the bar magnet and the external magnetic field,θ= 30°
Torque is related to magnetic moment (M) as:
T= MB sin θ
Hence, the magnetic moment of the magnet is 0.36 J T-1.
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 3: A short bar magnet placed with its axis at 30º with a uniform externalmagnetic field of 0.25 T ....
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