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Asked by rbhatt16 19th June 2018, 11:02 PM
Answered by Expert
Answer:
Let dn number of turns at a radius r in the spiral and having width dr. Then we have begin mathsize 12px style d n space equals space fraction numerator N over denominator open parentheses b minus a close parentheses end fraction d r end style
magnetic field dB at centre of spiral due to this dn turns is given by, begin mathsize 12px style d B space equals space fraction numerator mu subscript 0 cross times space d n cross times space i over denominator 2 space r end fraction space equals space fraction numerator mu subscript 0 space i over denominator 2 space r end fraction cross times fraction numerator N over denominator open parentheses b minus a close parentheses end fraction d r end style
magnetic field B due to entire spiral, begin mathsize 12px style B space equals space integral subscript a superscript b d B space equals space fraction numerator mu subscript 0 space i over denominator 2 end fraction cross times fraction numerator N over denominator open parentheses b minus a close parentheses end fraction integral subscript a superscript b fraction numerator d r over denominator r end fraction space equals space fraction numerator mu subscript 0 space end subscript i space N over denominator 2 space open parentheses b minus a close parentheses end fraction ln left parenthesis b divided by a right parenthesis end style 
 
Answered by Expert 20th June 2018, 11:53 AM
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