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.

Question

ROSHAN MUJEEB · Answer

nitially2T0=mg2T0=mg..(i) if&omega;&omega;be the frequency corresponding to the breaking of the string then current in the ring I=QT=Q2&pi;/&omega;=Q&omega;2&pi;I=QT=Q2&pi;/&omega;=Q&omega;2&pi; Magnetic moment of the loop M=iAM=iA The torque experinces now &tau;=MBsin90∘&tau;=MBsin90∘ ifT1T1andT2T2are the tensions in the string now, then (T1&minus;T2)d2=iAB(T1-T2)d2=iAB orT1&minus;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&minus;(iABd)T2=mg2-(iABd) AsT1>T2T1>T2 &there4;T1=3T02&there4;T1=3T02(Given) Hence 3T02=2T02+&omega;Q2&pi;d&times;&pi;R2&times;B3T02=2T02+&omega;Q2&pi;d&times;&pi;R2&times;B orT0=&omega;QBR2dT0=&omega;QBR2dor&omega;=dT0QBR2

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Magnetism

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship