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प्रश्न
A magnetised needle of magnetic moment 4.8 × 10−2 JT−1 is placed at 30° with the direction of uniform magnetic field of magnitude 3 × 10−2 T. Calculate the torque acting on the needle.
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उत्तर
The torque (τ) of a needle in uniform magnetic field will be given as
`vectau = vecM xx vecB =MB sintheta`
Here
M = magnetic moment = 4.8 ×10−2 J/T
B = magnetic field strength = 3 × 10−2 T
θ = angle with respect to field = 30°
So, the torque will be rewritten as
`vectau`= (4.8 ×10−2) × (3 × 10−2) sin 30°
`or vectau = 14.4 xx 10^-4 xx 1/2 = 7.2 xx 10^-4 N`
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