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q.

Asked by Arushi Juyal 13th December 2017, 7:24 AM
Answered by Expert
Answer:
Let us divide the top semi-cylindirical shell into small strips of width dw so that each divided strip behaves like wire of infinite length.
Let us assume the current i1 distributed uniformly and the current density α per unit width is given by 
 
begin mathsize 12px style alpha space equals space fraction numerator i subscript 1 over denominator pi r end fraction end style
magnetic field at center due to a single strip is begin mathsize 12px style fraction numerator mu subscript 0 space alpha space d w over denominator 2 pi r end fraction end style
Hence the magnetic fiel at center due to total current flowing in  top semi-cylindrical is begin mathsize 12px style integral fraction numerator mu subscript 0 space alpha space d w over denominator 2 pi r end fraction space equals space fraction numerator mu subscript 0 alpha over denominator 2 pi r end fraction pi r space equals space fraction numerator mu subscript 0 alpha over denominator 2 end fraction space equals space fraction numerator mu subscript 0 i subscript 1 over denominator 2 pi r end fraction end style
Since the current of same magnitude i1 flows but in opposite direction in bottom seni-cylindrical shell,
same magnetic field as calculated above is generated at center as calculated above.
 
Hence total field at center due to current flowing in both semi-cylindrical shell is begin mathsize 12px style B space equals space fraction numerator mu subscript 0 i subscript 1 over denominator pi r end fraction end style
Force per unit length on wire is B×i2×1 = begin mathsize 12px style fraction numerator mu subscript 0 i subscript 1 i subscript 2 over denominator pi r end fraction end style
I did not get any answer given in the options, but i believe this is the way to solve this problem
Answered by Expert 15th December 2017, 4:18 PM
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