# 0.4 Magnetic field due to current in a circular wire  (Page 3/5)

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Problem : Calculate magnetic field at the center O for the current flowing through wire segment as shown in the figure. Here, current through wire is 10 A and radius of the circular part is 0.1 m.

Solution : Magnetic field at O is contributed by long straight wire and circular wire. The direction of magnetic field at O due to straight part of the wire is into the plane of drawing as obtained by applying Right hand thumb rule for straight wire. The direction of current in the circular part is anticlockwise and hence magnetic field due to this part is out of the plane of drawing as obtained by applying Right hand thumb rule for circular wire.

The magnitude of magnetic field due to circular wire is :

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

The magnitude of magnetic field due to straight wire is :

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

Hence, magnitude of magnetic field at O is algebraic sum of two magnetic fields (we consider outward direction as positive) :

$B={B}_{C}-{B}_{S}=\frac{{\mu }_{0}I}{2R}-\frac{{\mu }_{0}I}{2\pi R}$

Putting values :

$⇒B=\frac{4\pi {10}^{-7}X10}{2X0.1}-\frac{4\pi {10}^{-7}X10}{2\pi X0.1}$ $⇒B=62.9X{10}^{-6}-20X{10}^{-6}=42.9\phantom{\rule{1em}{0ex}}\mu T$

The net magnetic field is acting out of the plane of paper.

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

The magnitude of magnetic field due to current in arc shaped wire can be obtained by integrating Biot-Savart expression in an appropriate range. Now, the integral set up for current in circular wire is :

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

Circular arc is generally referred in terms of the angle θ, it subtends at the center of the circle. From geometry, we know that :

$đl=Rđ\theta$ Substituting in the integral and taking the constant R out of the integral, we have :

$B=\frac{{\mu }_{0}I}{4\pi R}\int đ\theta$ $⇒B=\frac{{\mu }_{0}I\theta }{4\pi R}$

This is the expression for the magnitude of magnetic field due to current in an arc which subtends an angle θ at the center. Note that the expression is true for the circle for which θ = 2π and magnetic field is :

$⇒B=\frac{{\mu }_{0}IX2\pi }{4\pi R}=\frac{{\mu }_{0}I}{2R}$

Problem : Find the magnetic field at the corner O due to current in the wire as shown in the figure. Here, radius of curvature is 0.1 m for the quarter circle arc and current is 10 A.

Solution :

Here the straight line wire segment AB and CD when extended meet at O. As such, there is no magnetic field due to current in these segments. The magnetic field at O is, therefore, solely due to magnetic field due to quarter arc AC. The arc subtends an angle π/2 at its center i.e O. Now,

$B=\frac{{\mu }_{0}I\theta }{4\pi R}$

Putting values, we have :

$⇒B=\frac{{10}^{-7}X10X\pi }{0.1X2}=0.157X{10}^{-6}=0.157\phantom{\rule{1em}{0ex}}\mu T$

Since current in the arc is anticlockwise, magnetic field is perpendicular and out of the plane of drawing.

## Current in straight wire .vs. current in circular wire

A length of wire, say L, is given and it is asked to maximize magnetic field in a region due to a current I in the wire. Which configuration would we consider – a straight wire or a circular wire? Let us examine the magnetic fields produced by these two configurations.

If we bend the wire in the circle, then the radius of the circle is :

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