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**Hint:**In a current carrying coil the electric current creates a magnetic field which is more concentrated in the center of the coil than outside the loop. According to Biot-Savart’s law, the magnetic field at a point due to an element of a conductor carrying current is,

-Directly proportional to the strength of the current.

-Directly proportional to the length of the element.

-Directly proportional to the Sine of the angle between the element and the line joining the element to the point.

-Inversely proportional to the square of the distance between the element and the point.

**Complete step by step answer:**

Let us assume that the radius of the coil be R.

Then the magnetic field at distance r on the axis of coil is given by,

$B = \dfrac{{{\mu _o}I{R^2}}}{{2{{({r^2} + {R^2})}^{\dfrac{3}{2}}}}}$

Now when $r > > R$

$B = \dfrac{{{\mu _o}I{R^2}}}{{2{r^3}}}$

Since radius of the coil and current are constant we can say that,

$B \propto \dfrac{1}{{{r^3}}}......(1)$

Now at the center of the coil we have,

$r = R$

Hence, the expression for magnetic field becomes

$

B = \dfrac{{{\mu _o}I{r^2}}}{{{r^3}}} \\

\Rightarrow B = \dfrac{{{\mu _o}I}}{r} \\ $

Since, current is constant Therefore, we have

$B \propto \dfrac{1}{r}......(2)$

From (1) and (2) we can observe that the plot for magnetic field due to current carrying coil is

\[\begin{array}{*{20}{c}}

B& \propto &{\left\{ {\begin{array}{*{20}{c}}

{\dfrac{1}{{{r^3}}}}&{r > > R} \\

{\dfrac{1}{r}}&{r = R}

\end{array}} \right\}}

\end{array}\]

**It is clear from the above equations that option B is the correct representation.**

**Note:**When a current flows in a wire, it creates a circular magnetic field around the wire. This magnetic field can deflect the needle of a magnetic compass. The strength of a magnetic field is directly proportional to the current flowing. Therefore, if an alternating current is flowing, a magnetic field around the conductor will be produced, that is in phase with the alternating current.

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