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Home / Doubts and Solutions/JEE/Physics

Question

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Asked by rbhatt16 26th June 2018, 6:08 PM
Answered by Expert
Answer:
Let I be the current passing through the conductor that has the radius of cross section R as shown in figure.
Current density J = begin mathsize 12px style fraction numerator I over denominator pi space R squared end fraction end style
Let r be radius of half of cross section. Current i through half of cross section is given by,    begin mathsize 12px style i space equals space pi space r squared cross times J space equals space pi space r squared space cross times space fraction numerator I over denominator pi space R squared end fraction space equals space I cross times r squared over R squared space end style...........(1)
using eqn.(1) and using Ampher law, magnetic field B is given by, B = begin mathsize 12px style fraction numerator mu subscript 0 i over denominator 2 pi space r end fraction space equals space fraction numerator begin display style mu subscript 0 end style over denominator begin display style 2 pi space r end style end fraction cross times space I space r squared over R squared space equals space fraction numerator begin display style mu subscript 0 space I end style over denominator begin display style 2 pi space R squared end style end fraction space r space end style ........(2)
magnetic flux = magnetic field × Area = begin mathsize 12px style fraction numerator mu subscript 0 space I over denominator 2 space pi space R squared space end fraction integral subscript 0 superscript R r space d r space space equals space fraction numerator mu subscript 0 space I over denominator 4 space pi space space end fraction space equals space 10 to the power of negative 7 end exponent cross times space 10 space equals space 10 to the power of negative 6 end exponent space W b end style

Answered by Expert 29th June 2018, 11:52 AM
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