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The coercivity of a magnet is 5 103 A/m. The amount of current required to be passed in a solenoid of length 30 cm and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is A.
10
Coercivity = magnetic intensity H needed to demagnetise the magnet.
Inside a solenoid: H = nI, n = Nl = 1500.3 = 500 m-1
I = Hn = 5 103500 = 10 A
Answer: 10
A circular coil having 200 turns, 2.5 10-4 m2 area and carrying 100 current is placed in a uniform magnetic field of 1 T. Initially the magnetic dipole moment ( M) was directed along B. Amount of work, required to rotate the coil through 90 from its initial orientation such that M becomes perpendicular to B, is J.
The magnetic moment of a bar magnet is 0.5 Am2. It is suspended in a uniform magnetic field of 8 10-2 T. The work done in rotating it from its most stable to most unstable position is:
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