A thin uniform ring of mass m and electric charge q uniformly distributed rotates around an axis perpendicular to its plane and going through its centre. The angular momentum of the ring is 7.5 × 10–4 kg-m2/s. The ring is in a homogeneous magnetic field of field strength of 0.1 t and the lines of the magnetic induction are parallel with the plane of the ring. Torque exerted on the ring is? [the specific charge (charge-mass ratio) of the ring is [q/m=10−5c/kg][q/m=10−5c/kg].
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The torque on the ring is 3.75×10^−10 N.m
Explanation:
Q/m = 10^−5
Q =10^−5 m
Angular acceleration of ring = 7.5×10−4
Angular momentum = mass×area / time=7.5×10−4
Torque = NBIA / 2
= BQA / 2T
= 0.1 × Q × L^2 /2t
= 0.1 × 10^−5 × 7.5×10^−4 / 2
= 3.75×10^−10 N.m
Hence the torque on the ring is 3.75×10^−10 N.m
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