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(a) Write an expression of magnetic moment associated with a current $(I)$ carrying circular coil of radius $r$ having $N$ turns.

(b) 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 $(x,0,0)$

Answers (1)

a) $M=NIA=NI\pi r^2$

(b)

Due to the current element $dl$ considered at A the magnetic field at P is-

$dB=\frac{\mu _{0}I}{4\pi }\frac{\left | \vec{dl}\times \vec{r} \right |}{r^{3}}$

Since, $dl\times r=dlr$

$dB=\frac{\mu _{0}I}{4\pi }\frac{{dl} }{r^{2}}$

$dB=\frac{\mu _{0}I}{4\pi }\frac{dl}{(X^{2}+R^{2})}$

Consider a current element opposite to A, that is at B, then we can see that the y components of the magnetic, field due to this current element cancel and X component is only present.

So we can say that the net magnetic field is along the X-direction.

$dB_{X}=dB\cos \theta$

The net magnetic field at P is

$B=\int dB_{X}=\int dB\cos \theta$

$=\int \frac{\mu _{0}}{4\pi }\frac{Idl}{(X^{2}+R^{2})}\cos \theta$

$=\int \frac{\mu _{0}}{4\pi }\frac{Idl}{(X^{2}+R^{2})}\frac{R}{(X^{2}+R^{2})^{\frac{1}{2}}}$

$=\frac{\mu _{0}I\; R}{4\pi \; (X^{2}+R^{2})^{\frac{3}{2}}}\int dl=\frac{\mu _{0}IR\; 2\pi R}{4\pi (X^{2}+R^{2})^{\frac{3}{2}}}$

$B=\frac{\mu _{0}IR^{2}}{2(X^{2}+R^{2})^{\frac{3}{2}}}$

Posted by

Safeer PP

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