Magnetism Question 29
Question: A short bar magnet with its north pole facing north forms a neutral point at P in the horizontal plane. If the magnet is rotated by 90o in the horizontal plane, the net magnetic induction at P is (Horizontal component of earth’s magnetic field = $ B _{H} $ )
[EAMCET (Engg.) 2000]
Options:
A) 0
B) 2 BH
C) $ \frac{\sqrt{5}}{2}B _{H} $
D) $ \sqrt{5},B _{H} $
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Answer:
Correct Answer: D
Solution:
Initially Neutral point obtained on equatorial line and at neutral point $ | B _{H} |=| B _{e} | $ where BH = Horizontal component of earth’s magnetic field, Be = Magnetic field due to bar magnet on it’s equatorial line Finally Point P comes on axial line of the magnet and at P, net magnetic field $ B=\sqrt{B _{a}^{2}+B _{H}^{2}} $
$ =\sqrt{{{(2B _{e})}^{2}}+{{(B _{H})}^{2}}}=\sqrt{{{(2B _{H})}^{2}}+B _{H}^{2}}=\sqrt{5}\ B _{H} $
BETA
