Question

What is the expression of mutual inductance for shown concentric co – planar circular and regular hexagon $\left( {a > > r} \right)$ ?

Answer 1

Hint: Construct an equilateral triangle of height $d$ inside the hexagon. Find the magnetic field for one side of the hexagon and then for all of its sides. Use the Biot – Savart’s Law for this which can be expressed as:
$dB = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Idl.\sin \theta }}{{{r^2}}}$
where, $I$ is the amount of current in the loop and $r$ is the radius of that loop.

Complete step by step solution:
As this question is to be solved by using the Biot – Savart’s Law so, we should know what it is? The equation which gives the magnetic field produced due to the current carrying segment is known as Biot – Savart’s Law. The formula of Biot – Savart’s Law is:
 $dB = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Idl.\sin \theta }}{{{r^2}}}$
where, $I$ is the amount of current in the loop and $r$ is the radius of that loop.
Now, we have to make some construction in the figure given in the question. Therefore,

Let the current flowing in the hexagon be $I$.
Now, by using the Biot – Savart’s Law, the magnetic field at centre $O$ due to one side of hexagon is:
$
{B_1} = \dfrac{{{\mu _0}I}}{{2\pi d}}\cos {60^ \circ } \\
\Rightarrow {B_1} = \dfrac{{{\mu _0}I}}{{4\pi \dfrac{{\sqrt 3 a}}{2}}} \\
$
There are 6 sides in the hexagon, therefore, to calculate the magnetic field at centre $O$ due to all of its six sides we have to multiply ${B_1}$ with 6. So, -
$
\Rightarrow B = 6{B_1} \\
\Rightarrow B = 6\dfrac{{{\mu _0}I}}{{4\pi \dfrac{{\sqrt 3 }}{2}a}} \\
\therefore B = \dfrac{{\sqrt 3 {\mu _0}I}}{{\pi a}} \\
$
Hence, the flux through the circular loop can be calculated as –
$\phi = B \times \pi {r^2}$
Putting the value of magnetic field in the above equation of flux –
$
\phi = \dfrac{{\sqrt 3 {\mu _0}I}}{{\pi a}} \times \pi {r^2} \\
 \Rightarrow \phi = \dfrac{{\sqrt 3 {\mu _0}I{r^2}}}{a} \\
 $
Now, we know the relation between the mutual induction and flux which can be expressed as:
$\phi = MI$
where, $M$ is the mutual induction and $I$ is the current flowing in the loop.
So, the above relation can also be written as –
$ \Rightarrow M = \dfrac{\phi }{I}$
Putting the value of flux in the above expression of mutual induction, we get –
$
\Rightarrow M = \dfrac{{\dfrac{{\sqrt 3 {\mu _0}I{r^2}}}{a}}}{I} \\
\therefore M = \dfrac{{\sqrt 3 {\mu _0}{r^2}}}{a} \\
$
Hence, the required expression of mutual induction is $\dfrac{{\sqrt 3 {\mu _0}{r^2}}}{a}$.

Note: It is not necessary to find the magnetic field at the centre due to each side of the hexagon, you can find it from one side and multiply it with six to get the overall magnetic field of the loop.