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# Magnetic Field Due To Circular Current Loop - Practice Questions & MCQ

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

## Quick Facts

• Magnetic Field due to circular current loop is considered one of the most asked concept.

• 78 Questions around this concept.

## Solve by difficulty

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

Two concentric coils each of radius equal to  cm are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in weber/m2  at the center of the coils will be

If in a circular coil $A$ of radius$R$, current $I$ is flowing and in another coil $B$ of radius $2R$, a current $2I$ is flowing, then the ratio of the magnetic fields at the centre, $B_{A}$ and $B_{B}$ , produced by them will be

## Concepts Covered - 1

Magnetic Field due to circular current loop

Magnetic Field due to circular current loop at its centre:

Magnetic Field due to circular coil at Centre-

Consider a circular coil of radius a and carrying current  I  in the direction shown in Figure. Suppose the loop lies in the plane of the paper. It is desired to find the magnetic field at the centre  O of the coil. Suppose the entire circular coil is divided into a large number of current elements, each of length dl.

According to Biot-Savart law, the magnetic field  $d\vec{B}$ at the centre  O of the coil due to the current element  $I \overrightarrow{dl}$  is given by,
$\overrightarrow{\mathrm{dB}}=\frac{\mu_{0} \mathrm{I}(\overrightarrow{\mathrm{di}} \times \overrightarrow{\mathrm{r}})}{4 \pi \mathrm{r}^{3}}$

where $\vec{r}$ is the position vector of point O from the current element.The magnitude of  $\overrightarrow{\mathrm{dB}}$  at the centre  O  is

$\mathrm{dB}=\frac{\mu_{0} \text { Idlrsin } \theta}{4 \pi \mathrm{r}^{3}}$
$\therefore \mathrm{dB}=\frac{\mu_{0} \mathrm{Id} l \sin \theta}{4 \pi \mathrm{r}^{2}}$

The direction of  $\overrightarrow{\mathrm{dB}}$  is perpendicular to the plane of the coil and is directed inwards. Since each current element contributes to the magnetic field in the same direction, the total magnetic field  B  at the centre O  can be found by integrating the above equation around the loop i.e.

$\therefore \mathrm{B}=\int \mathrm{dB}=\int \frac{\mu_{0} \mathrm{Id}l \sin \theta }{4 \pi \mathrm{r}^{2}}$

For each current element, angle between $\overrightarrow{\mathrm{dI}}$ and $\vec{r}$ is $90^{\circ}$. Also distance of each current element from the center O  is a.

$\therefore B=\frac{\mu_{0} I \sin 90^{\circ}}{4 \pi \mathrm{r}^{2}} \int \mathrm{d} l$

But  $\int \mathrm{dl}=2 \pi \mathrm{r}= \text{total length of the coil}$

\begin{aligned} &\therefore B=\frac{\mu_{0} I}{4 \pi \mathrm{r}^{2}} 2 \pi \mathrm{r} \\ &\therefore B=\frac{\mu_{0} I}{2 \mathrm{r}} \end{aligned}

For N turns,

$B_0=B_{Centre}=\frac{\mu_{0}}{4\pi }\frac{2\pi Ni}{r}=\frac{\mu_{0} Ni}{2r}$

where N=number of turns, i= current and r=radius of a circular coil.

Magnetic field due to a current-carrying circular arc

Case 1:  Arc subtends angle theta at the centre as shown below then  $B_0=\frac{\mu_{0}}{4\pi } \frac{i \theta }{r}$

Proof:

Consider length element dl lying always perpendicular to $\vec{r}$.

Using the Biot-Savart law, the magnetic field produced at O is:

\begin{aligned} &\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \frac{I d \vec{l} \times \vec{r}}{r^{3}} \\ &d B=\frac{\mu_{0}}{4 \pi} \frac{I d l r \sin 90^{\circ}}{r^{3}} =\frac{\mu_{0}}{4 \pi} \frac{ I dl}{r^{2}} \ldots .(1) \end{aligned}

Equation (1) gives the magnitude of the field. The direction of the field is given by the right-hand rule. Thus, the direction of each of the dB is into the plane of the paper. The total field at O is

The angle subtended by element $dl$ is $d \theta$ at pt. O, therefore $dl= r d \theta$
\begin{aligned} &\mathrm{B}=\int \mathrm{dB}=\frac{\mu_{0}}{4 \pi} I \int_{\mathrm{0}}^{\theta} \frac{\mathrm{d} l}{\mathrm{r}^{2}} \\ &B=\frac{\mu_{0}}{4 \pi} I \int_{\mathrm{0}}^{\theta} \frac{\mathrm{rd} \theta}{\mathrm{r}^{2}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}} \theta \ldots . .(2) \end{aligned}

where the angle $\theta$ is in radians.

Case 2:  Arc subtends angle ($2\pi-\theta$) at the centre then $B_0=\frac{\mu_{0}}{4\pi }.\frac{(2\pi -\theta)i}{r}$

Case 3:The magnetic field of the Semicircular arc at the centre  is $B_0= \frac{\mu_{o}}{4\pi} \:\frac{\pi i}{r}= \frac{\mu_{o}i}{4r}$

Case 4: Magnetic field due to three-quarter Semicircular Current-Carrying arc at the centre $B_0=\frac{\mu_{o}}{4\pi}\:\frac{(2\pi-\frac{\pi}{2})i}{r}$

Special cases

1. If the  Distribution of current across the diameter then $B_0=0$

2. If Current between any two points on the circumference then $B_0=0$

3. Concentric co-planar circular loops carrying the same current in the Same Direction-

$B_{centre}=\frac{\mu_{o}}{4\pi}\:\left (2\pi i \right ) \left [\frac{1}{r_{1}}+\frac{1}{r_{2}} \right ]$

If the direction of currents are the same in concentric circles but have a different number of turns then

$B_{centre}=\frac{\mu_{o}}{4\pi}\:\left (2\pi i \right ) \left [\frac{n_1}{r_{1}}+\frac{n_2}{r_{2}} \right ]$

4. Concentric co-planar circular loops carrying the same current in the opposite Direction

$B_{centre}=\frac{\mu_{o}}{4\pi}\:\left (2\pi i \right ) \left [\frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]$

If the number of turns is not the same i.e $n_1\neq n_2$

$B_{centre}=\frac{\mu_{o}}{4\pi}\:\left (2\pi i \right ) \left [\frac{n_1}{r_{1}}-\frac{n_2}{r_{2}} \right ]$

5.  Concentric loops but their planes are perpendicular to each other

Then $B_{net}=\sqrt{{B_{1}}^2+{B_2}^2}$

6. Concentric loops but their planes are at an angle ϴ with each other

$B_{net}=\sqrt{{B_{1}}^2+{B_2}^2+2B_1B_2\cos\theta }$

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Magnetic Field due to circular current loop

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