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Home / Doubts and Solutions/JEE/Physics

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Asked by rbhatt16 19th July 2018, 9:32 AM
Answered by Expert
Answer:
 
Magnetic field at dB due to the current element idl is given by   begin mathsize 12px style fraction numerator mu subscript 0 over denominator 4 pi end fraction fraction numerator i space stack d l space with rightwards arrow on top space cross times space r with rightwards arrow on top over denominator r cubed end fraction end style .
Where r is the distance between P and current element dl as shown in figure.
The vectors and angle between them is shown in figure.
 
Hence magnitude of dB = begin mathsize 12px style fraction numerator mu subscript o i cross times r cross times d l over denominator r cubed end fraction sin left parenthesis 90 minus theta right parenthesis space end style
if we substitute dl = r×dθ, and r = a/cosθ in the above equation we get
begin mathsize 12px style open vertical bar d B close vertical bar space equals space fraction numerator mu subscript o space i space cos squared theta space d theta over denominator 4 pi space a end fraction end style
Answered by Expert 28th July 2018, 11:27 AM
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