# 0.4 Magnetic field due to current in a circular wire  (Page 2/5)

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## Current in circular wire and magnet

The directional attributes of the magnetic field due to current in circular wire have an important deduction. If the current in a circular loop is anticlockwise when we look from one end (face), then the same current is clockwise when we look from opposite end (face). What it means that if direction of magnetic field is towards you from one face, then the direction of magnetic field is away from you from the other end and vice versa.

The magnetic lines of force enters from the face in which current is clockwise and exits from the face in which current is anticlockwise. This is exactly the configuration with real magnet. The anticlockwise face of the circular wire is equivalent to north pole and clockwise face is equivalent to south pole of the physical magnet. For this reason, a current in a circular wire is approximately equivalent to a tiny bar magnet.

## Magnitude of magnetic field due to current in circular wire

Evaluation of Biot-Savart expression at the center of circle for current in circular wire is greatly simplified. There are threefold reasons :

1: The directions of magnetic fields due to all current elements at the center are same just as in the case of straight wire.

2: The linear distance (r) between current length element (d l ) and the point of observation (center of circular wire) is same for all current elements.

3: The angle between current length element vector (d l ) and displacement vector ( R ) is right angle for all current elements. Recall that angle between tangent and radius of a circle is right angle at all positions on the perimeter of a circle.

The magnitude of magnetic field due to a current element according to Biot-Savart law is given by :

$\mathrm{đB}=\frac{{\mu }_{0}}{4\pi }\frac{Iđl\mathrm{sin}\theta }{{r}^{2}}$

But, θ=90° and sin90°=1. Also, r = R = Radius of circular wire.

$⇒\mathrm{đB}=\frac{{\mu }_{0}}{4\pi }\frac{Iđl}{{R}^{2}}$

All parameters except "đl" in the right hand expression of the equation are constants and as such they can be taken out of the integral.

$B=\int \mathrm{đB}=\frac{{\mu }_{0}I}{4\pi {R}^{2}}\int đl$

The integration of dl over the complete circle is equal to its perimeter i.e. 2πR.

$⇒B=\frac{{\mu }_{0}I}{4\pi {R}^{2}}X2\pi R=\frac{{\mu }_{0}I}{2R}$

If the wire is a coil having N circular turns, then magnetic filed at the center of coil is reinforced N times :

$B=\frac{{\mu }_{0}NI}{2R}$

Problem : A thin ring of radius “R” has uniform distribution of charge, q, on it. The ring is made to rotate at an angular velocity “ω” about an axis passing through its center and perpendicular to its plane. Determine the magnitude of magnetic field at the center.

Solution : A charged ring rotating at constant angular velocity is equivalent to a steady current in circular wire. We need to determine this current in order to calculate magnetic field. For this, let us concentrate at any cross section of the ring. All the charge passes through this cross section in one time period of revolution. Thus, equivalent current is :

$I=\frac{q}{T}=\frac{q\omega }{2\pi }$

Now, magnetic field due to steady current in circular wire is :

$B=\frac{{\mu }_{0}I}{2R}$

Substituting for current, we have :

$⇒B=\frac{{\mu }_{0}q\omega }{4\pi R}$

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