A long wire is first bent into a circular coil of one turn and then into a circular coil of smaller radius having n turns. If the same current passes in both the cases, find the ratio of the magnetic fields produced at the centres in the two cases.

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Solution

Case 1:
Number of turns =1
Current = I
Length of wire = l
radius $= \frac{l}{2 π} = r_{1}$ (1)
Case 2:
Number of turns = n
Current = I
Length = l
radius $= \frac{l}{2 π n} = r_{2}$
Using relation
$B = \frac{μ_{0} n I}{2 r}$
$\frac{B_{1}}{B_{2}} = \frac{μ_{0} I}{2 r_{1}} \times \frac{2 r_{2}}{μ_{0} n I}$
$= \frac{r_{2}}{n r_{1}} = \frac{l}{2 π n} \times \frac{1}{n} \times \frac{2 π}{l} = \frac{1}{n^{2}}$

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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 turns. If the same current passes in both the cases, find the ratio of the magnetic fields produced at the centres in the two cases.

Case 1:Number of turns =1Current = ILength of wire = lradius=l2π=r1 (1)Case 2:Number of turns = nCurrent = ILength = lradius=l2πn=r2Using relationB=μ0nI2rB1B2=μ0I2r1×2r2μ0nI=r2nr1=l2πn×1n×2πl=1n2