A magnetic moment $20 A - m^{2}$ is freely suspended is a uniform magnetic field of strength $6 T$ . Work done to rotate it by $60^{0}$ will be

70 J

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60 J

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80 J

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90 J

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Solution

The correct option is B $60 J$
$t = \to M \times \to B = M B s i n θ$
$\int d w = \int t d θ$
$W = \int_{θ_{i}}^{θ_{f}} M B s i n θ d θ$
$= M B (c o s θ_{i} - c o s θ_{f})$
$θ_{i} = 0^{\circ}$
$O_{f} = 60^{\circ}$
$= \frac{M B}{2}$
$= \frac{20 \times 6}{2}$
$= 60 J$

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A magnetic moment 20A−m2 is freely suspended is a uniform magnetic field of strength 6T. Work done to rotate it by 600 will be

The correct option is B 60Jt=→M×→B=MBsinθ∫dw=∫tdθW=∫θfθiMBsinθdθ=MB(cosθi−cosθf)θi=0∘Of=60∘=MB2=20×62=60J

A magnetic moment $20 A - m^{2}$ is freely suspended is a uniform magnetic field of strength $6 T$ . Work done to rotate it by $60^{0}$ will be

The correct option is B $60 J$
$t = \to M \times \to B = M B s i n θ$
$\int d w = \int t d θ$
$W = \int_{θ_{i}}^{θ_{f}} M B s i n θ d θ$
$= M B (c o s θ_{i} - c o s θ_{f})$
$θ_{i} = 0^{\circ}$
$O_{f} = 60^{\circ}$
$= \frac{M B}{2}$
$= \frac{20 \times 6}{2}$
$= 60 J$