Question

Write an expression of magnetic moment associated with a
current (I) carrying circular coil of radius r having N turns.
Consider the above mentioned coil placed in YZ plane with its
centre at the origin. Derive expression for the value of magnetic
field due to it at point (x, 0, 0).​

Answer 1

I wrote this as my answer

Answer 2

Given :

Radius of the current carrying coil placed in YZ plane = r

No of turns in the coil = N

The centre of the coil is at the origin

To Derive :

Expression of magnetic moment associated with the coil .

And expression for the value of the magnetic field due to the coil at a point ( x, 0 , 0 )

Solution :

We know that for a current carrying coil magnetic moment m is given as :

$m = n \times I \times A$

Here n is no of turns , I is current and A is area .

So, $m = N I \pi r^2$

Here r is radius of the coil

Now according to  Biot Savart's law we have :

$B =\frac{ \frac{\mu_0 I}{4\pi}\times I \times ( \vec {dl}\times \vec{R})}{R^3}$

Here dl is current element and R is position vector .

Or,$dB = \frac{\mu_o .I.dl.R. sin\theta}{4\pi R^3}$

Here , $\theta$ is angle between dl and R , which  is equal to 90°, so :

$dB = \frac{\mu_0I. dl}{4 \pi R^2}$

Since the y component of magnetic field at the (x,0,0) gets cancelled , so:

$B= \int(dB)_x$

Now ,

$(dB)_x = \frac{\mu_0Idl \times cos\phi}{4\pi R^2}$

Here $\phi$ is angle between X-axis and dB.

So, $(dB)_x= \frac{\mu_0Idl}{4\pi R^2}\times \frac{r}{R}$

               $= \frac{\mu_0 I dl r }{4\pi R^3}$

Now , since ;

 $R =\sqrt{x^2+r^2}$

So, $(dB)_x= \frac{\mu_0 I dl r }{4\pi (\sqrt{x^2 + r^2 })^3}$

                $= \frac{\mu_0 Idl r }{4 \pi (x^2 + r^2)^{\frac{3}{2} }}$

∴$B = \int (dB)_x$

     $= \int \frac{\mu_0Idl r }{4\pi(r^2+x^2)^{\frac{3}{2} }}$

    $= \frac{\mu_0 I r }{4\pi (r^2 + x^2 )^\frac{3}{2} } \times \int \int\limits^{2\pi r}_0 {} \, dl$

    $=\frac{\mu_0 I r }{4 \pi (x^2+ r^2)^{\frac{3}{2}}} \times 2 \pi r$

    $=\frac{\mu_0 I r^2 }{2 (x^2+ r^2)^{\frac{3}{2}}}$