A small loop of area of cross-section 10-4 m2 is lying concentrically and coplanar inside a bigger loop of radius 0.628 m. A current of 10 A is passed in the bigger loop. The smaller loop is rotated about its diameter with an angular velocity ω. The magnetic flux linked with the small loop will be

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10-9 sinωt

10-9 cosωt

10-10 sinωt

10-10 cosωt

answer is B.

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Detailed Solution

We know that Φ=BAcos⁡ωt Here B=μ0i2r ∴Φ=μ0i2r×Acos⁡ωt =4×3⋅14×10−7102×0⋅628×10−4×cos⁡ωt =10−9cos⁡ωt

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