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A small square loop of wire of side lis placed inside a large square loop of wire of side L ( > > l). The loops are coplanar and their centres coincide. What is the mutual inductance of the system? |
Answer» Solution :CONSIDERING the large loop to be made up of four rods each of length L, the FIELD at the centre, i.e., at a distance (L/2) from each rod, will be `B= 4 xx (mu_0 )/(4pi) I/d [ sin alpha + sin beta]` i.e., `B = 4 xx (mu_0)/(4pi) (l)/((L//2)) xx 2 sin 45` `i.e., B_1 = (mu_0 )/(4pi) (8sqrt2)/(L) l` So the flux linked with smaller loop `phi_2 =B_1 S_2 = (mu_0)/(4pi) (8 SQRT2I)/(L)l^2` and hence `M = (phi_2)/(I)= 2sqrt2 (mu_0)/(pi) (l^2)/(L) rArr M prop (l^2)/(L)` |
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