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A long wire carries a steady curent . It...

A long wire carries a steady curent . It is bent into a circle of one turn and the magnetic field at the centre of the coil is `B`. It is then bent into a circular loop of `n` turns. The magnetic field at the centre of the coil will be

Text Solution

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Let R be the radius of one turn circular coil. Then,
`B_0=(mu_0)/(4pi)(2piI)/(R)`
If r is the radius of circular coil of n turns, then
`2pirn=2piR` or `r=R//n`
So, `B=(mu_0)/(4pi)(2pinI)/(r)=(mu_0)/(4pi)(2pinI)/((R//n))=(mu_0)/(4pi)(2piI)/(R)n^2`
`=n^2B_0`
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