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

The magnetic field on the axis of the circular current carrying 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 $d \bar{\ell}$ element that is diametrically opposite to previously $d \ell$ element cause $d \vec{B}^{\prime}$.
Due to symmetry the components of $d \vec{B}$ and $d \vec{B}^{\prime}$ perpendicular to the x-axis cancel each other. i.e., these components add to zero.
The x-components of the $d \vec{B}$ 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 magnetic field due to current element is

$
\mathrm{dB}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi} \int \frac{\mathrm{dl} \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 \sin \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 \cos \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} \ldots}
\end{aligned}
$

Total magnetic field along axis $=B_{\text {axis }}=\int d B_x=\int d B \sin \theta$

$
\because \int d B_y=\int d B \cos \theta=0
$

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., $\int d \ell=2 \pi R$
So, we get

$
\Rightarrow B_{\text {axis }}=\frac{\mu_0 I R^2}{2\left(x^2+R^2\right)^{3 / 2}} \text { (on the axis of a circular loop) }
$
 

• - If $\mathrm{x}>\mathrm{R}$, then $B=\frac{\mu_0 I R^2}{2 x^3}$.
  - At centre, $x=0 \Rightarrow B_{\text {centre }}=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi \mathrm{~N} i}{R}=\frac{\mu_0 N i}{2 R}=B_{\max }$