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Q.

The pole strength of a bar magnet is 48 Am and the distance between its poles is 25cm. The moment of the couple by which it can be placed at an angle of 300 with the uniform magnetic field of flux density 0.15 N/Am will be.

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a

12Nm

b

18Nm

c

0.9Nm

d

0.5Nm

answer is C.

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Detailed Solution

Moment of the couple means torque and flux density means magnetic field.∴Torque, Γ=MBSinθ=m l BSinθ      [m=pole strength]=48 x 0.25 x 0.15 x 12= 0.9 Nm
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