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Question: A bar magnet of length 10 cm and having the pole strength equal to 10<sup>–3</sup> weber is kept in ...

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 (2) equal to 4π×10−34 \pi \times 10 ^ { - 3 }Tesla. It makes an angle of 30o with the direction of magnetic induction. The value of the torque acting on the magnet is

A

2π×10−7N×m2 \pi \times 10 ^ { - 7 } N \times m

B

2π×10−5N×m2 \pi \times 10 ^ { - 5 } N \times m

C

0.5N×m0.5 N \times m

D

0.5×102N×m0.5 \times 10 ^ { 2 } N \times m ()

Answer

2π×10−7N×m2 \pi \times 10 ^ { - 7 } N \times m

Explanation

Solution

Torque τ=MBHsin⁡θ\tau = M B _ { H } \sin \theta(1) τ=MBHsin⁡θ\tau = M B _ { H } \sin \theta

=0.1×10−3×4π×10−3×sin⁡30o=10−7×4π×12= 0.1 \times 10 ^ { - 3 } \times 4 \pi \times 10 ^ { - 3 } \times \sin 30 ^ { o } = 10 ^ { - 7 } \times 4 \pi \times \frac { 1 } { 2 }

=2π×10−7N×m= 2 \pi \times 10 ^ { - 7 } N \times m