Question

Obtain an expression for magnetic flux density B at the centre of a circular coil of radius R, having N turns and carrying a current $I$.

Answer 1

Magnetic field at O due to small element $\delta l$

$\delta B=\frac{{\mu }_0}{4\pi }\frac{i\delta lsin\theta }{R^2}$

$\theta ={90}^o$

$\therefore \delta B=\frac{{\mu }_0}{4\pi }\frac{i\delta l}{R^2}$

$\overrightarrow{B}$at O due to the entire coil.

$B=\int dB=\int \frac{{\mu }_0}{4\pi }\frac{idl}{R^2}$

$=\frac{{\mu }_0}{4\pi }\frac{i}{R^2}\cdot 2\pi R=\frac{{\mu }_{0}i}{2R}$

If there are N tums then $B=\frac{{\mu }_{0}Ni}{2R}$

Putting the value of dB from eq. (i) and si n $\varphi =a/r$ (from fig.) we get

Now since the terms ${\mu }_0/4\pi$, i and a are constant and r has the same value foe all the elements of loop. Thus the term ${\mu }_{0}ia/4\pi r^3$ can be factored out of integration.

$\therefore B=\frac{{\mu }_0}{4\pi }\frac{ia}{r^3}\int dl$.

But $\int dl=2\pi a$ (cirumference of the loop) and $r=(a^2+x^2{)}^{1/2}$.

$\therefore B=\frac{{\mu }_0}{4\pi }\frac{2\pi i{a}^2}{(a^2+x^2{)}^{3/2}}=\frac{{\mu }_{0}i{a}^2}{2(a^2+x^2{)}^{3/2}}$

It is a coil of N turns, thn each turn will contribute equally to B. Thus,

$\therefore B=\frac{{\mu }_0}{4\pi }\frac{2N\pi i{a}^2}{(a^2+x^2{)}^{3/2}}=\frac{{\mu }_{0}Ni{a}^2}{2(a^2+x^2{)}^{3/2}}$

The direction of magnetic $\overrightarrow{B}$ is aong the axis of the loop and of the coil.

Magnetic field at the centre : At the centre of the coil, we have.

$\therefore B=\frac{{\mu }_0}{4\pi }\frac{2N\pi i}{a}=\frac{{\mu }_{0}Ni}{2a}$