Question

A circular coil of closely wound $N$ turns and radius $r$ carries a current $I$ . Write the expression for the following:
(i) The magnetic field at its centre
(ii) The magnetic moment of this coil

Answer 1

Use the Biot-Savart law to calculate the magnetic field. Make necessary substitutions in the law and finally integrate it to get the answer. After finding out the magnetic field, the magnetic moment can be calculated using a simple formula. Recall the formula and the law. Keep in mind the limits needed for integration will be for the whole circumference of the ring.

Formula used:
The magnetic field for a small current element is given by $d\vec B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{i(d\vec l \times \vec r)}}{{{r^2}}}$
The magnetic moment is given by $M = NiA$
Where in above the symbols have their usual meanings.

Complete step by step solution:
i) Given,
Radius of ring $= r$
Current flowing $= I$
Number of turns $= N$
We know that
$d\vec B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{i(d\vec l \times \vec r)}}{{{r^3}}}$
For a small element $dl$ on the ring, the current is given by $i = NI$ since there are $N$ turns in the ring.
Also the distance between the element $dl$ and the point where the magnetic field is to be calculated becomes the radius of the ring, $r$ and the angle between the element $dl$ and distance vector is ${90^ \circ }$ .
Putting these in the above equation we get,
$\Rightarrow dB = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{NIdlr \times \sin {{90}^ \circ }}}{{{r^3}}}$
$\Rightarrow dB = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{NIdl}}{{{r^2}}}$
Therefore integrating both sides we get,
$\Rightarrow B = \int\limits_0^{2\pi r} {\dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{NIdl}}{{{r^2}}}}$
$\Rightarrow B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{NI}}{{{r^2}}}\int\limits_0^{2\pi r} {dl}$
The integration of $dl$ is $l$
Putting this is the above equation,
$\Rightarrow B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{NI}}{{{r^2}}}[l]_0^{2\pi r}$
Putting the appropriate limits, we get
$\Rightarrow B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{NI}}{{{r^2}}}2\pi r$
On simplifying we get
$\Rightarrow B = \dfrac{{{\mu _o}NI}}{{2\ r}}$
Which is the required magnetic field at the centre of the ring.
ii)Now, the magnetic moment of the circular ring is given by
$M = NiA$
$\therefore M = NI\pi {r^2}$
Which is the required answer.

Note: The formula used here $d\vec B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{i(d\vec l \times \vec r)}}{{{r^2}}}$ is known as the Biot Savart’s Law. It can be used to find out the magnetic field of any combination by suitable integration. Remember the magnetic field at standard positions for other arrangements also.