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A long wire is first bent into a circular coil of one turn and then into a circular coil of smaller radius, having n identical turns. If the same current passes in both the cases, find the ratio of the magnetric fields produced at the centre in the two cases. |
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Answer» Solution :Let a wire of length L is first bent in a single CIRCULAR loop of radius `R_1 = (L)/(2pi)` and a CURRENT I passes through it. Then the magnetic field at the centre of loop : `B_1= (mu_0 I)/(2 R_1) = (mu_0 I)/(2L) cdot 2pi "" ……..(i)` When the same wire is bent into a circular COIL of n turns then radius of coil `R_2 = L/(2 pi R)` . Now on passing same current I, the field at the centre point will be `B_2 = (mu_0 nI)/(2 R_2) = (mu_0 n I)/(2L) cdot 2 pi r = (mu_0 n^2 I)/(2L) cdot 2 pi "".............(ii)` `IMPLIES (B_1)/(B_2) = 1/(n^2) "or" B_2 = n^2B_1`. |
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