The inner loop has an area of $5 \times 10^{- 4} m^{2}$ and a resistance of $2 Ω$ (figure). The larger circular loop is fixed and has a radius of 0.1 m. Both the loops are concentric and coplanar. The smaller loop is rotated with an angular velocity $ω r a d s^{- 1}$ about its diameter. The magnetic flux with the smaller loop is

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$π \times 10^{- 9} \cos ω t W b$

$2 π \times 10^{- 6} W b$

$π \times 10^{- 9} W b$

Zero

answer is C.

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Detailed Solution

Magnetic field due to larger loop

$= \frac{μ_{0} I}{2 r} = \frac{4 π \times 10^{- 7} \times 1}{2 \times 0.1} T = 2 π \times 10^{- 6} T$

Now, magnetic flux linked with the smaller loop, $ϕ$

$= N B A \cos ω t = 1 \times 2 π \times 10^{- 6} \times 5 \times 10^{- 4} \cos ω t$

$= π \times 10^{- 9} \cos ω t w e b e r$

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