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5.Magnetism and Matter
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A bar magnet of length $10 \,cm$ and having the pole strength equal to $10^{-3}$ weber is kept in a magnetic field having magnetic induction $ (B)$ equal to $4\pi \times {10^{ - 3}}$ Tesla. It makes an angle of $30°$ with the direction of magnetic induction. The value of the torque acting on the magnet is
A
$2\pi \times {10^{ - 7}}\,N \times m$
B
$2\pi \times {10^{ - 5}}\,N \times m$
C
$0.5\,N \times m$
D
$0.5 \times {10^2}\,N \times m$
Solution
(a)Torque $\tau = M{B_H}\sin \theta $
$ = 0.1 \times {10^{ – 3}} \times 4\pi \times {10^{ – 3}} \times \sin {30^o} = {10^{ – 7}} \times 4\pi \times \frac{1}{2}$
$ = 2\pi \times {10^{ – 7}}\,N \times m$
Standard 12
Physics
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