A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 2.50 S.I. unit and R = 1m)
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answer is 00002.50.
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Detailed Solution
dϕ= (2πxdx)kxt2ϕ=23kt2πx3 dϕdt=4kπtx33 ⇒τ=23ktx2QπR2(2πxdx)xE2πx=4πktx33; E=23ktx2; 43ktQR2R55∫dτ=43ktQR2∫0R x4dx At t=15sec,τ=2.50Nm