Question

A long wire carries a steady current. When it is bent in a circular form the magnetic field at its centre is B. Now if this wire is bent in a circular loop of n turns what is the magnetic field at its centre?
a).$nB$
b).${{n}^{2}}B$
c).$2nB$
d).$2{{n}^{2}}B$

Answer 1

We have to apply the relation for Magnetic field at the centre of the current carrying circular loop of radius r. If the wire contains n turns then the magnetic field at centre of the current carrying loop will become n times the initial value of the magnetic field. If any change in radius and current will occur, the magnitude of the magnetic field will change.

Complete answer:
Let us assume the length of long wire is $L$ and a steady current $I$is flowing in the wire.
Now this wire is bent in a circular loop, let us assume the radius of the circular loop is$r$ .
Suppose Magnetic field at centre of Current Carrying Circular loop $=B$
Then Length of Wire = Circumference of Circular loop
$L=2\pi r$
$r=\dfrac{L}{2\pi }$---- (i)
Since, Magnetic field at centre of Circular loop is given by the relation,
$B=\dfrac{{{\mu }_{0}}I}{2r}$
Put value of $r$ from Equation 1, so value of Magnetic field becomes,
B=\dfrac{{{\mu }_{0}}I}{2{}^{L}/{}_{2\pi }}$$$$=\dfrac{{{\mu }_{0}}I\pi }{L}----- (ii)
Now it is given in the question that the wire is now bent into $n$ circular turns. If the same wire is turned into $n$ circular turns then the radius of wire becomes less.
Let new radius of wire is {{r}^{'}}$$$$({{r}^{'}}Then Length of Wire = Circumference of Circular loop \[L=n(2\pi {{r}^{'}})
${{r}^{'}}=\dfrac{L}{2n\pi }$----- (iii)
So, Magnetic field at centre of Circular loop for $n$ turns is given by the relation,
${{B}^{'}}=\dfrac{n{{\mu }_{0}}I}{2{{r}^{'}}}$
Put the value of ${{r}^{'}}$ from equation (iii), we get
{{B}^{'}}=\dfrac{n{{\mu }_{0}}I}{2{}^{L}/{}_{2\pi n}}$$$$=\dfrac{{{n}^{2}}{{\mu }_{0}}I\pi }{L}----- (iv)
Compare Equation (ii) and (iv), we get
${{B}^{'}}={{n}^{2}}B$

So option b) is the correct answer.

Additional information:
Magnetic field at axial point due to circular current carrying loop is given by $B=\dfrac{{{\mu }_{0}}I{{r}^{2}}}{2{{({{a}^{2}}+{{r}^{2}})}^{3/2}}}$where r is the radius of circular loop and a is the distance from centre of circle to axial point.

Note:
We have to remember that whenever we have to find a magnetic field at the centre of the current carrying loop we always have to take axial distance zero. When many turns of current carrying loop are present then magnitude of magnetic field is obtained by multiplying the value that number of times.