Magnetic Field Due To Circular Current Loop MCQ - Practice Questions with Answers

Magnetic Field due to circular current loop at its centre: 

Magnetic Field due to circular coil at Centre-

$$
B_0=B_{\text {Centre }}=\frac{\mu_0}{4 \pi} \frac{2 \pi N i}{r}=\frac{\mu_0 N i}{2 r}
$$

where $\mathrm{N}=$ number of turns, $\mathrm{i}=$ current and $\mathrm{r}=$ radius of a circular coil.
Similarly, if Arc subtend angle theta at the centre as shown below then

$$
B_0=\frac{\mu_0}{4 \pi} \cdot \frac{\theta i}{r}
$$
 

 

similarly, if Arc subtends angle $(2 \pi-\theta)$ at the centre then

$$
B_0=\frac{\mu_0}{4 \pi} \cdot \frac{(2 \pi-\theta) i}{r}
$$
 

So the magnetic field of Semicircular arc at the centre is

$$
B_0=\frac{\mu_o}{4 \pi} \frac{\pi i}{r}=\frac{\mu_o i}{4 r}
$$
 

So Magnetic field due to three quarter Semicircular Current-Carrying arc at the centre is

$$
B_0=\frac{\mu_o}{4 \pi} \frac{\left(2 \pi-\frac{\pi}{2}\right) i}{r}
$$
 

• The direction of the magnetic field-

Right-Hand Thumb Rule gives the direction of the magnetic field of Circular Currents.

Right-hand thumb rule is stated below:

If the fingers are curled along the current, the stretched thumb will point towards the magnetic field.

1. If the current is in a clockwise direction then the direction of the magnetic field is away from the observer or perpendicular inwards.

2. If the current is in anti-clockwise direction then the direction of the magnetic field is towards the observer or perpendicular outwards

• Special cases-

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

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 having 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 $B_{\text {net }}=\sqrt{B_1^2+B_2^2}$

 

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

$B_{\text {net }}=\sqrt{B_1^2+B_2^2+2 B_1 B_2 \cos \theta}$