Magnetism Question 185
Question: The small magnets each of magnetic moment 10 A-m2 are placed end-on position 0.1m apart from their centres. The force acting between them is
[MNR 1994]
Options:
A) $ 0.6\times 10^{7}N $
B) $ 0.06\times 10^{7}N $
C) $ 0.6N $
D) $ 0.06N $
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Answer:
Correct Answer: C
Solution:
$ F=\frac{{\mu _{0}}}{4\pi }( \frac{6M{M}’}{d^{4}} ) $ in end-on position between two small magnets.
$ \
Therefore \ F={{10}^{-7}}( \frac{6\times 10\times 10}{{{( 0.1 )}^{4}}} )=0.6N $
BETA
