A bar magnet of moment $4 A m^{2}$ is placed in a non-uniform magnetic field. If the field strength at poles are 0.2 T and 0.22 T then the maximum couple acting on it is

0.04Nm

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0.84Nm

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0.4 Nm

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0.44Nm

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Solution

The correct option is B 0.84Nm
Average of the two field strengths at the poles is $\frac{0.2 + 0.22}{2} = 0.21$
Copuling( torque): $M B = 4 \times 0.21 N m = 0.84 N m$

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