# derive an expression for magnetic field on the axis of a circular current loop

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## Derive the expression for magnetic field at a point on the axis of a circular current carrying loop.

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Updated on : 2022-09-05

In the above diagram,Solution Verified by Toppr

Let O be the centre of the ring, R be the radius of the ring. The point P lie outside the ring on the axis, let the distance between the P and o be x, where we have to calculate the magnetic field.

Let dB

be the magnetic field due to a small length dl of the ring,

As per bio-severt rule,

dB = 4πr 2 μ o Idl We know that r= (x 2 +R 2 ) So db = 4π x 2 +R 2 μ o Idl

This field have vertical

dB y = dB

sinθ and horizontal component

dB x = dB cosθ,

Vertical component will cancel out each other, only horizontal components are responsible,

So, B = dB 1 + dB 2 +..... dB = 4π R 2 +x 2 μ o Idlsinθ dB = 4π(R 2 +x 2 ) 2 3 μ o Ix

The magnetic field due to the circular current loop of radius a at a point which is a distance R away, and is on its axis,

So B = 2(R 2 +x 2 ) 2 3 μ o Ix 2

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## Magnetic Field on the Axis of a Circular Current Loop

Let us test the direction of the magnetic field produced by a current-carrying portion of the wire. Also, learn the derivation of magnetic field on the axis of a circular current loop in detail, at BYJU’S - The Learning App.

PhysicsElectricity And MagnetismMagnetic Field On The Axis Of A Circular Current Loop

## Magnetic Field on the Axis of a Circular Current Loop

We know that there exists a relationship between electricity and magnetism. The field around the magnet generates a magnetic field, and the rotating magnets in a generator produce electricity. The phenomenon which relates electricity and magnetism is known as the electromagnetic force. In this article, let us learn about the magnetic field on the axis of a circular current loop in detail.

**Table of Contents:**

Magnetic Field on the Axis of a Circular Current Loop – Derivation

Practice Problem

Frequently Asked Questions – FAQs

## Magnetic Field on the Axis of a Circular Current Loop – Derivation

Before we know more about the magnetic field on the axis of a circular current loop, let us understand the basic law of magnetism. We know that there exists an equation that describes the magnetic field generated by constant current. The Biot–Savart law correlates the magnitude of the magnetic field with the length, proximity, and direction of the electric current. The Biot–Savart law is an equation that gives the magnetic field produced by a current-carrying segment. The current-carrying element is considered a vector quantity. Biot–Savart law is given by the equation:

|dB|=(μ04π)(Idlsinθr2)———–(1)

Where

μ0 is the permeability of free space and is equal to 4π × 10-7 TmA-1.

Now let us learn the derivation of a magnetic field on the axis of a circular current loop. Let us evaluate the magnetic field due to a circular coil along its axis. The evaluation requires summing up the effect of minute current elements “idl”. “i” is the steady current and the evaluation is carried out in a vacuum or free space.

Refer to the image given below. From the figure, we can see that “i” represents a circular loop carrying a steady current. From the figure, we can observe that the circular loop is placed in the y-z plane with its centre at the origin O and has a radius “R”. The x-axis is the axis of the loop.

x = distance P from the centre O of the loop.

dl= a conducting element of the loop.

The Biot-Savart law gives the magnitude of the magnetic field (dB) due to current element (dl)

|dB|=(μ04π)(I|dl∗r|r3)

Now r2=x2+R2———–(2)

Any element of the loop will be perpendicular to the displacement vector from the element to the axial point.

For example, the element dl present in the figure is in the y-z plane. Whereas, the displacement vector r from dl to the axial point P is in the x-y plane.

Hence, |dl∗r|=r.dl Thus,

|dB|=(μ04π)Idl(x2+R2)

The direction of dB is as shown in the figure and is perpendicular to the plane formed by dl and r. It has an x-component dBx and a component perpendicular to x-axis, dB⊥

A null result is obtained when the components perpendicular to the x-axis are summed over, and they cancel out.

The dB⊥ component due to dl is cancelled by the contribution due to the diametrically opposite dl element. This is represented in the above figure.

Hence, only the x-component survives. The net contribution along the x-direction can be obtained by integrating dBx = dBcosθ over the loop.

cosθ=R(x2+R2)12———–(3)

From equation (2) and (3)

dBx=μ0Idl4πR(x2+R2)32

The summation of elements dl over the loop yields 2πR, the circumference of the loop. Thus, the magnetic field at P due to the entire circular loop is

B=Bxi^=μ0IR22(x2+R2)32i^

When x=0, we may obtain the field at the center of the loop

B0=μ0I2Ri^

Since the circular wire forms a closed loop, the magnetic field lines are produced as shown in the figure below.

The direction of the magnetic field created is given by the right-hand thumb rule.

Accelerate Your JEE Main & Advanced Exam Preparation byWatching this Video about Magnetism and Force on Dielectric in Charged Capacitor

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### Practice Problem

1. A circular coil of wire has 100 turns of radius 12 cm, and carries a current of 1 A in a clockwise direction when viewed from the right side. Find the magnitude and direction of the magnetic field:

i) at the centre of the coil, and

Given: N = 100, r = 0.12m, i = 1 A

Magnetic field at the centre of the circular coil is given by:

B=μ0NI2R

=4π∗10−7∗100∗12∗0.12

=5.23∗10−4T Related links

Polarisation By Scattering

Murphy’s law Types of rocks Atmosphere

## To Learn about Magnetic Fields Due to Circular Loops, Click on the Video Below

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## Frequently Asked Questions – FAQs

### State right-hand thumb rule.

The right-hand thumb rule states that “Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field”.

### The equation that gives the magnetic field produced by a current-carrying segment.

## Derive the expression for magnetic field at a point on the axis of a circular current loop.

Derive the expression for magnetic field at a point on the axis of a circular current loop.

Derive the expression for magnetic field at a point on the axis of a circular current loop.

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