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Asked by MUKUL | 04 Oct, 2019, 19:42: PM
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EMF induced in the loop , E = - (dΦ/dt) = - A (dB/dt)  ......................(1)
 
where Φ is magnetic flux. Magnetic flux = B×A , where B is magnetic flux density and A is area of loop.
 
Magnetic Flux density B at centre of loop due to current Io passing through wire is given by,
 
begin mathsize 14px style B space equals space fraction numerator mu subscript o I over denominator 2 pi open parentheses a divided by 2 close parentheses e end fraction space equals space fraction numerator mu subscript o I over denominator pi space a space e end fraction end style ...........................(2)
using eqn.(1) and (2), EMF induced in loop,
 
begin mathsize 14px style E space equals space A space fraction numerator d B over denominator d t end fraction space equals space space l left parenthesis e minus 1 right parenthesis a fraction numerator mu subscript o over denominator pi space a space e end fraction fraction numerator d I subscript o over denominator d t end fraction end style
Current I in the loop is obtained by dividing the induced emf by resistance of the loop and it is given by,
 
begin mathsize 14px style I space equals space l left parenthesis e minus 1 right parenthesis space fraction numerator mu subscript o over denominator pi space space e space R end fraction fraction numerator d I subscript o over denominator d t end fraction end style
Force between the induced current and current passing through long wire is given by
 
begin mathsize 14px style F space equals space fraction numerator mu subscript o I subscript o I over denominator 2 pi a end fraction l space space equals space fraction numerator mu subscript o I subscript o over denominator 2 pi a end fraction space l squared space left parenthesis e minus 1 right parenthesis space fraction numerator mu subscript o over denominator pi space space e space R end fraction fraction numerator d I subscript o over denominator d t end fraction space equals space fraction numerator mu subscript o superscript 2 space I subscript o superscript 2 space l squared open parentheses e space minus space 1 close parentheses over denominator 2 pi squared space a space e space R space d t end fraction end style
In the above equation to get Force, dIo is considered as dIo = ( Io - 0 ) = Io
 
Acceleration a' is obtained by dividing force F by mass m
 
a' = begin mathsize 14px style F over m space equals space fraction numerator mu subscript o superscript 2 space I subscript o superscript 2 space l squared open parentheses e space minus space 1 close parentheses over denominator 2 pi squared space m space a space e space R space d t end fraction end style
velocity v of loop after time dt staring from rest is  given by
 
begin mathsize 14px style v space equals space a apostrophe cross times d t space space equals space fraction numerator mu subscript o superscript 2 space I subscript o superscript 2 space l squared open parentheses e space minus space 1 close parentheses over denominator 2 pi squared space m space a space e space R space end fraction end style
 
 
Answered by Thiyagarajan K | 04 Oct, 2019, 21:37: PM

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