A long insulated copper wire is closely wound as a spiral of 'N' turns. The spiral has an inner radius ‘a' and an outer radius ′b′. The spiral lies in the X−Y plane and a steady current 'I' flows through the wire. The Z-component of the magnetic field at the center of the spiral is :

Hint: In this question, first we assume an elementary ring of radius r and thickness ‘dr’ and then find the number of turns in this elementary ring . After this we will find the magnetic field at the centre O. To find the magnetic field due to the whole spiral, we will integrate it from a to b.

Complete answer:
Let us consider an elementary ring of radius r and thickness dr in which current I is flowing.



Number of turns in this elementary ring dN =$\dfrac{N}{{b - a}}dr$
We know that magnetic field at the centre of a ring is given by:
B=$\dfrac{{{\mu _0}NI}}{{2r}}$
Where,
${\mu _0}$ is permeability of free space.
N is the number of terms.
I is currently flowing.
r is the radius of the ring.


Thus magnetic field at the centre O due to this ring dB=$\dfrac{{{\mu _0}dNI}}{{2r}}$
Putting the value of dN, we get:
dB=$\dfrac{{{\mu _0}NIdr}}{{2(b - a)r}}$
To get the net magnetic field at the centre of the spiral , we will integrate the above equation from ‘a’ to ‘b’.
$\int\limits_0^B {dB} $ =$\int\limits_a^b {\dfrac{{{\mu _0}NIdr}}{{2(b - a)r}}} = \dfrac{{{\mu _0}NI}}{{2(b - a)}}\int\limits_a^b {\dfrac{{dr}}{r}} = \dfrac{{{\mu _0}NI}}{{2(b - a)}}\left[ {\ln r} \right]_a^b$
$ \Rightarrow B = \dfrac{{{\mu _0}NI}}{{2(b - a)}}\ln \dfrac{b}{a}$
Therefore, the required magnetic field is $ \Rightarrow B = \dfrac{{{\mu _0}NI}}{{2(b - a)}}\ln \dfrac{b}{a}$

Note:
In this question, you should know how to calculate the magnetic field in the centre of the circular wire. The direction of the magnetic field in this case is given by right hand thumb rule. You should know the Biot-savart law which is used to calculate the magnetic field. The magnitude of magnetic field is given by:
$dB = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Idl\sin \theta }}{{{r^2}}}$