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Q.

A conductor is bent in the form of concentric semicircles as shown in the figure. The magnetic field at the point common center O is

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a

zero

b

μ0I6a

c

μ0Ia

d

μ0I4a

answer is B.

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Detailed Solution

magnetic field for a semicircle B=μ0i4r here B=μoi4a1-12+14-18+116-132......... B= μoi4a1+14+116+164......-μoi4a21+14+116+164........ B =μoi4a43-μoi8a43=μoi6a
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