A bar magnet length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the center of the magnet. If BH=0.4 G, the magnetic moment of the magnet is (1 G=10 -4T)

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2.880×103JT−1

2.880×102JT−1

2.880JT−1

28.80JT−1

answer is C.

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Detailed Solution

Magnetic field of dipole is balanced by horizontal component of earth's magnetic field.2Bsin⁡θ=BH⇒2μ04πmr20.07r=0.4×10−4 here r=0.072+0.182=0.193m⇒210−7m0.03730.070.193=0.4×10−4⇒m=20.57Am Magnetic Moment =m(2L)=20.57×0.14=2.88Am2 or 2.88JT−1

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