A small square loop of wire of side l is placed inside a large square loop of wire of side L (L > l). The loop are coplanar and their centre coincide. The mutual inductance of the system is proportional to
A
l/L
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B
l2/L
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C
L/l
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D
L2/l
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Solution
The correct option is Bl2/L Magnetic field produced due to large loop B=μ04π8√2iL
Flux linked with smaller loop ϕ=B(l2)=μ04π8πil2L ∴ϕ=Mi⇒M=πi=μ04π.8√2l2L⇒M∝l2L