Magnetic Field On The Axis Of Circular Current Loop MCQ - Practice Questions with Answers

Magnetic field on the axis of circular current loop: 

In the figure, it is shown that a circular loop of radius R carrying a current $I$. Application of Biot-Savart law to a current element of length $dl$ at angular position $\theta$ with the axis of the coil.
the current in the segment $d\ell$ causes the field $d\overline{B}$ which lies in the x -y plane as shown below.
Another symetric $d{\overline{\ell }}^{\prime }$ element that is diametrically opposite to previous $d\ell$ element cause $d\overline{B}$.
Due to symmetry, the components of $d\overline{B}$ perpendicular to the x -axis cancel each other. I.e these components add to zero.
The x-components of the $d\overrightarrow{B}$ 's combine to give the total field $\overrightarrow{B}$ at point P .

We can use the law of Biot-Savart to find the magnetic field at a point P on the axis of the loop, which is at a distance x from the center.

$d\overline{\ell }$ and $\hat{r}$ are perpendicular and the direction of field $d\overline{B}$ caused by this particular element $d\overline{\ell }$ lies in the x-y plane.
The net magnetic field is $\mathbf{B}=\frac{{\mu }_0\mathrm{I}}{4\pi }\int \frac{\mathrm{d}\mathbf{l}\times \hat{\mathbf{r}}}{{\mathrm{r}}^2}$.
Since $r^2=x^2+R^2$
the magnitude $dB$ of the field due to element $d\overline{\ell }$ is:

$$dB=\frac{{\mu }_{0}I}{4\pi }\frac{d\ell }{(x^2+R^2)}$$

The components of the vector $dB$ are

$$\begin{array}{rl}& d{B}_x=dB\cos \theta =\frac{{\mu }_{0}I}{4\pi }\frac{d\ell }{(x^2+R^2)}\frac{R}{{(x^2+R^2)}^{1/2}}\dots \\ & d{B}_y=dB\sin \theta =\frac{{\mu }_{0}I}{4\pi }\frac{d\ell }{(x^2+R^2)}\frac{x}{{(x^2+R^2)}^{1/2}}\end{array}$$

To obtain the x-component of the total field $\overrightarrow{B}$, we integrate equation (1), including all the $d\overrightarrow{\ell }$ 's around the loop.
Everything in this expression except $d\overrightarrow{\ell }$ is constant and can be taken outside the integral.
The integral $d\overrightarrow{\ell }$ of is just the circumference of the circle, i.e.,

$$\int d\ell =2\pi a$$
 

So, we finally get
$B_{\text{axis }}=\frac{{\mu }_{0}I{R}^2}{2{(x^2+R^2)}^{3/2}}($ on the axis of a circular loop)
If $\mathrm{x}\gg \mathrm{R}$, then $B=\frac{{\mu }_{0}I{R}^2}{2x^3}$.

At centre,

$$x=0\Rightarrow B_{\text{centre }}=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi \mathrm{N}i}{R}=\frac{{\mu }_{0}Ni}{2R}=B_{max}$$