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

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 $I \overrightarrow{d l}$ is given by,

$
\overrightarrow{\mathrm{dB}}=\frac{\mu_0 \mathrm{I}\left(\overrightarrow{\mathrm{di}} \times \overrightarrow{\mathrm{r}^{\prime}}\right)}{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

$
\begin{aligned}
& \mathrm{dB}=\frac{\mu_0 \mathrm{Id} l \mathrm{rsin} \theta}{4 \pi \mathrm{r}^3} \\
& \therefore \mathrm{~dB}=\frac{\mu_0 \mathrm{Id} l \sin \theta}{4 \pi \mathrm{r}^2}
\end{aligned}
$

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{I} \mathrm{~d} l \sin \theta}{4 \pi \mathrm{r}^2}
$

For each current element, angle between \$loverrightarrow\{lmathrm\{d 1$\}\} \$ and $\$ \mid v e c\{r\} \$ is $\$ 90^{\wedge}\{\operatorname{lcirc}\} \$. Also distance of each current element from the center O is a.

$
\begin{aligned}
& \therefore B=\frac{\mu_0 I \sin 90^{\circ}}{4 \pi \mathrm{r}^2} \int \mathrm{~d} l \\
& \text { But } \int \mathrm{dl}=2 \pi \mathrm{r}=\text { total length of the coil } \\
& \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_{\text {Centre }}=\frac{\mu_0}{4 \pi} \frac{2 \pi N i}{r}=\frac{\mu_0 N i}{2 r}
$

where $\mathrm{N}=$ number of turns, $\mathrm{i}=$ current and $\mathrm{r}=$ 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 .

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 d l}{r^2} \ldots
\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 $d l$ is $d \theta$ at pt. O , therefore $d l=r d \theta$

$
\begin{aligned}
& \mathrm{B}=\int \mathrm{dB}=\frac{\mu_0}{4 \pi} I \int_0^\theta \frac{\mathrm{d} l}{\mathrm{r}^2} \\
& B=\frac{\mu_0}{4 \pi} I \int_0^\theta \frac{\mathrm{rd} \theta}{\mathrm{r}^2}=\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}} \theta \ldots . .
\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} \cdot \frac{(2 \pi-\theta) i}{r}
$
 

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

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

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_{\text {centre }}=\frac{\mu_o}{4 \pi}(2 \pi i)\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_{\text {centre }}=\frac{\mu_o}{4 \pi}(2 \pi i)\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_{\text {centre }}=\frac{\mu_o}{4 \pi}(2 \pi i)\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_{\text {centre }}=\frac{\mu_o}{4 \pi}(2 \pi i)\left[\frac{n_1}{r_1}-\frac{n_2}{r_2}\right]
$
 

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

Then $B_{n e t}=\sqrt{B_1{ }^2+B_2{ }^2}$

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

$B_{n e t}=\sqrt{B_1^2+B_2^2+2 B_1 B_2 \cos \theta}$