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A non-conducting ring of radius 'r' has charge 'Q'. A magnetic field perpendicular to the plane of th ring changes as dB/dt.the torque experienced by the ring is?

A non-conducting ring of radius 'r' has charge 'Q'. A magnetic field perpendicular to the plane of th ring changes as dB/dt.the torque experienced by the ring is?

Grade:12

1 Answers

Rathod Shankar AskiitiansExpert-IITB
69 Points
12 years ago

Dear M.vs vivek vardhan,

 

torque = u*B  (this is cross product)

 

 u = the magnetic dipole moment 

B =the magnetic field vector

u = I*A    = (current * area)

u x B = u*B*Sin(theta)     here theta is 90    and I=dQ/dt

 

Torque=  dQ/dt * B * (Π *r^2)  =   Q* dB/dt  * (Π *r^2)

 

 

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Rathod Shankar Singh

IIT Bombay

 

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