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

Options

A. `nB`
B. `n^(2)B`
C. `2nB`
D. `2n^(2)B`

Answer

2

Explanation

The magnetic field at the center of circular coil is
`B = (mu_(i))/(2r)`
where r = radius of circle `= (|)(2pi) (:' 1 =2pir)`
`:. B = (mu_(0)i)/(2) xx (2u)/(|) =(mu_(0)ipi)/(|)..(i) `
when wire of length bents into a circular loops of n turns then
` 1 = nxx 2 pir' implies r' (1)/(n xx 2pi)`
Thus , new magnetic field
`B' = (mu_0ni)/(2r') = (mu_(0)ni)/(2) xx (n xx 2pi)/(|)`
`= (mu_(0)ipi)/(|) xxn^(2)`
`=n^(2) B [f r oma eq (i)]` .