A uniform disc of radius r and mass m is charged uniformly with charge q This disc is placed flat on a rough horizontal surface having coefficient of friction on μ. A uniform magnetic field is present in a circular region (a>r) but varying at kt3 as shown in fig. Then the time (in second) after which the disc begin to rotate is 2x. Find x=? Given r=1m,m=18kg,q=(1/3)c,μ=0.1,k=4(S.I.units),g=10m/s2
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answer is 6.
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Detailed Solution
Induced Electric field ∫E→.dr→=dϕdt ⇒E2πx=πx2dBdt ⇒E=x2dBdt ⇒E=3kxt22For small elemental part dτ=3kxt22×q2πxdxπr2×x⇒τ=3kt2qr2∫0rx3dx=3kqt2r24 Torque due to friction force τ=∫0rdτ=∫0rμm2πxdxπr2gx=23μmgrFor rotational equilibrium, both torques balances each other.3kqt2r24=23μmgr ⇒t=12sec