A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field, B with the distance, d, from the centre of the conductor, is correctly represented by the figure :

CONCEPT:

Ampere circuital law- According to the ampere circuital law, we are having the relationships between the current and magnetic field produced by the conductor. It states that the line integral of the magnetic field intensity along the imaginary close path is equal to the product of the current enclosed by the path and permeability of the medium and it is written as;

\(\oint {\vec B.d\vec l = {\mu _o}I} \)

Here, B is the magnetic field, \(\mu_o\) is the permeability and 'I' is the current.

CALCULATION:

As we know from equation (1) we have;

\(\oint {\vec B.d\vec l = {\mu _o}I} \)     -----(1)

Now, let us consider a cylinder having a radius R which carries a current 'I'. 

Let us take three cases to get the plot of the magnetic field with distance d.

Case 1- Consider an imaginary loop of radius 'd'. Inside (d < R) Magnetic field inside the conductor.

\(dI'=\frac{I}{\pi R^2}\times \pi d^2\)

Now, on putting the value of dI' in the equation (1) we have;

\(\oint {\vec B_{inside}.d\vec l = {\mu _o}\frac{I}{\pi R^2}\times \pi d^2} \)

For the cylinder we have;

\(B_{inside}(2\pi d)={ {\mu _o}\frac{I}{\pi R^2}\times \pi d^2} \)

⇒ \(B_{inside} = \frac{{{\mu _0}\,i}}{{2\pi {R^2}}}d\)

Therefore the straight line passes through the origin

Case 2- At the surface (d = R)

Now by using equation (1) we have;

\(\oint {\vec B.d\vec l = {\mu _o}I} \)

\(B(2\pi d)={ {\mu _o}\frac{I}{\pi R^2}\times \pi d^2} \)

\(B= \frac{{{\mu _0}\,i}}{{2\pi \,R}}\)      ----(3)

It is maximum at the surface.

Case 3 - For the outside (d > R). Using equation (1) we have;

\(\oint {\vec B_{outside}.d\vec l = {\mu _o}I} \)

⇒ \(B_{outside} = \frac{{{\mu _0}\,i}}{{2\pi \,d}}\)

or B_outside ∝ \(\frac{1}{d}\)  (Hyperbolic)    ----(4) 

Therefore by combining the cases, we have a plot of the magnitude of the magnetic field, B with the distance, d as

Hence, option 3) is the correct answer.