A bar magnet of length 10 cm and having the pole strength equal to 10-3 Wb is kept in a magnetic field having magnetic induction equal to 4π x 10-3 T. It makes an angle 30o with the direction of magnetic induction. The value of the torque acting on the magnet is μ0=4π×10−7WbA−1m
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a
2π×10−7Nm
b
2π×10−5Nm
c
0.5 Nm
d
0.5×10−2Nm
answer is C.
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Detailed Solution
Pole strength = 10−3Wb=10−3μ0AmMagnetic moment, M = m(2l)∴M=10−3μ0×0⋅10=10−4μ0 Now τ=MBsinθ =10−4μ0×4π×10−3×sin30∘=10−44π×10−7×4π×10−3×12=12=0⋅5Nm