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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Magnetic field on the axis of circular current loop is considered one of the most asked concept.

  • 17 Questions around this concept.

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

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:

Concepts Covered - 1

Magnetic field on the axis of circular current loop

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\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\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_{\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 \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}}} ....(1)\\ \\ {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{array}

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_{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 }

 

 

 

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Magnetic field on the axis of circular current loop

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