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 $d l$ at angular position $\theta$ with the axis of the coil.
the current in the segment $d \ell$ causes the field $d \bar{B}$ which lies in the x -y plane as shown below.
Another symetric $d \bar{\ell}^{\prime}$ element that is diametrically opposite to previous $d \ell$ element cause $d \bar{B}$.
Due to symmetry, the components of $d \bar{B}$ perpendicular to the x -axis cancel each other. I.e these components add to zero.
The x-components of the $d \vec{B}$ 's combine to give the total field $\vec{B}$ 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.

$d \bar{\ell}$ and $\hat{r}$ are perpendicular and the direction of field $d \bar{B}$ caused by this particular element $d \bar{\ell}$ 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 $d B$ of the field due to element $d \bar{\ell}$ is:

$$
d B=\frac{\mu_0 I}{4 \pi} \frac{d \ell}{\left(x^2+R^2\right)}
$$

The components of the vector $d B$ are

$$
\begin{aligned}
& d B_x=d B \cos \theta=\frac{\mu_0 I}{4 \pi} \frac{d \ell}{\left(x^2+R^2\right)} \frac{R}{\left(x^2+R^2\right)^{1 / 2}} \ldots \\
& d B_y=d B \sin \theta=\frac{\mu_0 I}{4 \pi} \frac{d \ell}{\left(x^2+R^2\right)} \frac{x}{\left(x^2+R^2\right)^{1 / 2}}
\end{aligned}
$$

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

$$
\int d \ell=2 \pi a
$$
 

So, we finally get
$B_{\text {axis }}=\frac{\mu_0 I R^2}{2\left(x^2+R^2\right)^{3 / 2}}($ on the axis of a circular loop)
If $\mathrm{x} \gg \mathrm{R}$, then $B=\frac{\mu_0 I R^2}{2 x^3}$.

At centre,

$$
x=0 \Rightarrow B_{\text {centre }}=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi \mathrm{~N} i}{R}=\frac{\mu_0 N i}{2 R}=B_{\max }
$$