Question:
A small square loop of wire of side l is placed inside a large square loop of wire of side L.(L>>l).The loops are coplanar and their centres coincide.Find the mutual induction of the system.
Answer:
Let the current I pass through the outer coil of side L. Due to this the magnetic field at the centre of the coil will be B = µ0 I / L
As it is given that L >> l, we can consider the magnetic field B to be constant over the entire cross-sectional area of the inner coil of side l. Hence, flux associated with the small coil will be Ф1 = BA1 = B l2
Ф1 = (µ0 I / L) * l2
By the definition of mutual inductance Ф1 = M12 I2 = M12 I
Hence, M21 = M = M12 = Ф1 / I
= (µ0 I / L) * l2 * (1/I)
= µ0 l2 / L
The value of M shows that it is independent of the value of current and depends on the dimension of the two coils only.
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