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A charge Q = 1C is uniformly distributed on a ring which is free to rotate about a light horizontal rod of length 1m. The rod is suspended by light inextensible strings and a magnetic field B=10T is applied.The initial tentions in the strings are T=100N.If the breaking strength of the each string is 1.5T , find the maximum angular velocity with which the wheel can be rotated.


A charge Q = 1C is uniformly distributed on a ring which is free to rotate about a light horizontal rod of length 1m. The rod is suspended by light inextensible strings and a magnetic field B=10T is applied.The initial tentions in the strings are T=100N.If the breaking strength of the each string is 1.5T , find the maximum angular velocity with which the wheel can be rotated.


 


 


 


 


 



Grade:10

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
one year ago
nitially2T0=mg2T0=mg..(i)
ifωωbe the frequency corresponding to the breaking of the string then current in the ring
I=QT=Q2π/ω=Qω2πI=QT=Q2π/ω=Qω2π
Magnetic moment of the loop
M=iAM=iA
The torque experinces now
τ=MBsin90∘τ=MBsin90∘
ifT1T1andT2T2are the tensions in the string now, then
(T1−T2)d2=iAB(T1-T2)d2=iAB
orT1−T2=2iABd...(i)T1-T2=2iABd...(i)
alsoT1+T2=mg...(iii)T1+T2=mg...(iii)
Solving equations (ii) and (iii) we get
T1=mg2+(iABd)T1=mg2+(iABd)andT2=mg2−(iABd)T2=mg2-(iABd)
AsT1>T2T1>T2
∴T1=3T02∴T1=3T02(Given) Hence
3T02=2T02+ωQ2πd×πR2×B3T02=2T02+ωQ2πd×πR2×B
orT0=ωQBR2dT0=ωQBR2dorω=dT0QBR2

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