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

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Syllabus

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Magnetic field on the axis of circular current loop is considered one of the most asked concept.
23 Questions around this concept.

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EASY

A current $i$ ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is

infinite

zero

$\frac{\mu _{0}}{4\pi }.\frac{2i}{r}\, tesla$

$\frac{2i}{r}\, tesla$

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A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a/2 and 2a is:

0.25

0.5

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Concepts Covered - 1

Magnetic field on the axis of circular current loop

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 $dl$ 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 previously $d\ell$ element cause $d\vec {B'}$ .

Due to symmetry the components of $d\vec {B}$ and $d\vec {B'}$ 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 $\mathbf{dB}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi} \int \frac{\mathrm{d} \mathbf{l} \times \hat{\mathbf{r}}}{\mathrm{r}^{2}}$ .

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

$\begin{array}{l}{d B_{x}=d B sin \theta=\frac{\mu_{0} I}{4 \pi} \frac{d \ell}{\left(x^{2}+R^{2}\right)} \frac{R}{\left(x^{2}+R^{2}\right)^{1 / 2}}} ....(1)\\ \\ {d B_{y}=d B \cos \theta=\frac{\mu_{0} I}{4 \pi} \frac{d \ell}{\left(x^{2}+R^{2}\right)} \frac{x}{\left(x^{2}+R^{2}\right)^{1 / 2}}}....(2) \end{array}$

$\begin{aligned} &\text{Total magnetic field along axis}=B_{axis}= \int d B_{x}=\int d B \sin \theta \\ &\because \int d B_{y}=\int d B \cos \theta=0 \end{aligned}$

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
$\Rightarrow B_{axis}=\frac{\mu_{0} I R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} \text { (on the axis of a circular loop) }$

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