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Asked by MUKUL | 04 Oct, 2019, 07:42: PM

Expert Answer

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 I_o 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, dI_o is considered as dI_o = ( I_o - 0 ) = I_o

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, 09:37: PM

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