Question

A steady current flows in a long wire. It is bent into a circular loop of one turn and
the magnetic field at the centre of the coil is $B$. If the same wire is bent into a circular loop
of $n$ turns, the magnetic field at the centre of the coil is:
A. $\dfrac{B}{n}$
B. $nB$
C. $n{{B}^{2}}$
D. ${{n}^{2}}B$

Answer 1

Hint: When we bend a long wire to form a circular loop, the circumference of the loop is equal to the length of the wire. We will use the formula of magnetic field at the centre of the loop to find the ratio of magnetic field in the two cases.

Formula used:
$B=\dfrac{{{\mu }_{o}}I}{2R}$

Complete step by step answer:
When the wire is bent into a circular loop of one turn, the length of the wire will be equal to
the circumference of the loop
Let’s take the length of wire be $l$
Radius of loop = $R$
$2\pi R=l$
Magnetic field at the centre of the loop of radius $R$, $B=\dfrac{{{\mu }_{o}}I}{2R}$
Where $I$is the value of current passing through the loop
Now the same wire is bent into circular loop of $n$ turns
Therefore radius of the new loop will be,
$n\times 2\pi R'=2\pi R$
$R'=\dfrac{R}{n}$
Magnetic field at the centre of loop will be $B=\dfrac{n{{\mu }_{o}}I}{2R'}$
$\begin{aligned} & B=n\dfrac{{{\mu }_{o}}I}{2\dfrac{R}{n}}=\dfrac{{{n}^{2}}{{\mu }_{o}}I}{2R} \\\ & B={{n}^{2}}B \\\ \end{aligned}$

Value of magnetic field at the centre of the loop with $n$ turns will be ${{n}^{2}}$ times the
magnetic field at the centre of the loop with a single turn.
Hence, the correct option is D.

Note: While calculating the magnetic field through the loop of $n$ turns, remember that the current through the loop will become $n$ times of the current through a single turn. Also, we have ignored the thickness of the wire while calculating the radius of the coil with $n$ turns, as the wire is long and its thickness is negligible.