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In this article we are going to **derive an expression for magnetic field on the axis of circular current carrying loop**. So without talking so much let’s gets started…

Inside Story

# MAGNETIC FIELD ON THE AXIS OF A CIRCULAR CURRENT CARRYING LOOP

### AT ANY POINT ON THE AXIS

Let’s consider a circular loop of radius **a** with centre **C**. Let the plane of the coil be perpendicular to the plane of the paper and current be flowing in the direction shown.

Suppose, **P** is the any point on the axis at direction **r** from the centre. See figure below.

Now consider a **current** element on the top (**L**) where current comes out of the paper normally whereas at the bottom (**M**) enters into the plane of the paper normally.

According to the **Biot-Savart Law**, **magnetic field** due to **current** element at the point **P** is given as-

Where, **a** is the radius of the circular current carrying loop and **r** is the distance of the point **P** from the centre of the loop along the axis.

The direction of is perpendicular to **LP** and along the **PQ**. Similarly, the magnitude of the **magnetic field** can be obtained due to the current element at the bottom and the direction is along the **PQ**‘, where .

Now, resolving due to the current element at **L** and **M**. components balance each other and net magnitude of magnetic field is given as-

So after taking all the constant terms out from the integral, we get-

If loop contains **N** turns of coils then, the magnetic field is given as-

### AT THE CENTRE OF THE LOOP

If we place the point at the centre of the loop then it means value of distance of the point from the centre is zero i.e **r = 0**. Put **r = 0** in the above derived formula, then we get-

If the loop contains **N** turns of coils, it becomes-

Watch this video for more understanding.

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