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A small square loop of wire of side l is placed inside a large square loop of wire of side L (L gt gt l ). The loops are coplanar and their centres coincide. Find the mutual inductance of the system. |
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Answer» Solution :Let a current I pass through the LARGE square LOOP of SIDE L. Magnetic field at the centre O of the loop. B = 4 x magnetic field due to each side `therefore B=4xx(mu_0I)/(4PI(L/2)) [ sin 45^@+ sin 45^@]` `therefore B=4xx(mu_0I)/(4pixxL/2)[1/sqrt2+1/sqrt2]` `=(2mu_0I)/(piL)(1/sqrt2+1/sqrt2)` `therefore B=(2sqrt2mu_0I)/(piL)`....(1) Since l is very small compared to L, VALUE of B can be considered uniform over the area `A =pil^2`of the inner loop. `therefore` magnetic flux linked with small square loop `phi=AB` `therefore phi=l^2 xx (2sqrt2mu_0I)/(piL)`...(2) [`because` From equation (1)] Mutual inductance of the system of two loops `M=phi/I` `therefore M=(2sqrt2mu_0l^2)/(piL)` [`because` From equation (2)] `therefore M prop l^2/L` |
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