A short bar Magnet with its north pole facing North forms a neutral point P in the horizontal plane. If the magnet is rotated by 900 in the horizontal plane, the net magnetic induction at P is:
(Horizontal component of earths magnetic field is BH)
0
2BH
√52BH
√5BH
A
√52BH
B
0
C
2BH
D
√5BH
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Solution
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At P, the equatorial field due to magnet =BH. After rotation, P lies on the axis of the bar magnet.
Hence, at P, field due to magnet =2BH Net field is the vector sum of BH and 2BH, both at right angles to each other giving, Bnet=√5BH
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