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please answer this

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Asked by Prashant DIGHE 6th December 2019, 10:31 PM
Answered by Expert
Answer:
Figure shows a charged ring of radius R rotating with linear speed about axis OP.
Let a charge q is distributed uniformly, so that the ring has linear charge density λ per unit length.
 
Let us consider a small element of length dl at A. Electroc fileld dE at P due to this element of length dl is given by
 
dE = ( λ dl ) / [ (4πεo) 2R2 ]
Direction of this field is along the direction of line AP. The component resolved along the line OP is dE cos45 .
 
Hence electric field in the direction of line OP is given by,   dE = (1/√2 ) ( λ dl ) / [ (4πεo) 2R2 ] .............(1)
 
electric field E in the direction of line OP due to full ring is obtained by integrating eqn.(1) through whole length of ring
 
E = (1/√2 ) ( λ 2πR ) / [ (4πεo) 2R2 ] = [1/( 2√2 )] {  q  / [ (4πεo) 2R2 ] } ....................(2)
 
magnetic field at P due to rotation of charged ring is calculated as follows.
 
Equivalent current of rotating charged ring = charge passing through the cross section of wire per unit time
 
Equivalent current i = q / [ (2πR)/v ]  = λv.................(3)
 
Magnetic field dB by Biot_Severt's law at P due to line element dl  is given by,
 
dB = [ ( μo/4π) i dl sin 45 ]/ ( 2R2 ) ...........................(4)
 
Magenetic flux density due to whole ring is obtained by integrating eqn.(4)
 
B = (1/2√2) [ ( μo/4π) λv 2πR ]/ ( R2 ) = (1/2√2) [ ( μo/4π) q v ]/ ( R2 ) ..........................(5)
 
Substitution for current is done in the above equation using eqn.(3)
 
from eqn.(2) and (5) , we get the ratio of electric field to magnetic flux density is given by
 
( E / B ) = 1 / ( μo εo v )  = c2 / v 
Answered by Expert 8th December 2019, 2:08 PM
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