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Question

A circular conductor of radius ro=1cm has an internal field,


¯¯¯¯¯H=104ro(1a2sinaroroacosaro)ˆaϕ(A/m)

where, a=π2r0

A
8π Ampere
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B
8πAmpere
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C
4π Ampere
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D
4πAmpere
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Solution

The correct option is B 8πAmpere

According to Ampere’s law,

Ienc=r=r0H.dl
ro=102 m

2π0104r0(4r20π2sinπ22r20πcosπ2).r0dϕ

=1042π04r20π2dϕ

lenc=104.4r20π×2=8πAmpere


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