Question

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

• A. $nB$
• B. $n^{2}B$
• C. $3nB$
• D. $n^{3}B$

Answer 1

Answer: (B)

Suppose the length of wire is $l$
Then, for the circle of n turns
The magnetic field, $B=\frac{\mu _{0} I n}{2 R}$ , where R is the radius of the circle
Then $B_{s}=\frac{\mu _{0} \, I}{2 R}$ =B......(i)
Here, $2\pi R=l$ .......(ii)
Now, if the same wire is turned into a coil of n turns of radius a then,
$2\pi a\times n=l$ .......(iii)
Thus, $2\pi R=2\pi a\times n$
$\Rightarrow \, \, a=\frac{R}{n}$
Now, the magnetic field for the coil of n turns, is
$B_{n}=\frac{\mu _{0} I}{2 a}\times n$
$=\frac{\mu _{0} I}{2 R}\times n^{2}$ $\left(\because \, \, a = \frac{R}{n}\right)$
$=n^{2}B$