0.3 Magnetic field due to current in straight wire  (Page 4/4)

In order to nullify the magnetic field at D due to current in wire at A, the direction of magnetic field due to current in the wire at C should be equal and opposite. This means that the current in the wire at C should be opposite that of wire at A. Hence, the direction of current in the wire at C should be into the plane of drawing. Now, the magnitudes of magnetic fields due to currents are equal. Let the current in second wire be I, then:

Two straight wires carrying current

$$\frac{{\mu }_{0}I}{2\pi X5}=\frac{{\mu }_{0}X10}{2\pi X15}$$ $$\Rightarrow I=\frac{50}{15}=3.34A$$

Applying right hand rule for vector cross product, we realize that magnetic field due to each arm of the hexagon for given current direction (clockwise) is into the plane of hexagon. As such, we can algebraically add magnetic field due to each arm to obtain net magnetic field.

$$B=6B_a$$

where $B_a$ is magnetic field due to current in one of the arms. Now, we consider one of the arms of hexagon as shown in the figure. Here,

Magnetic field at the center due current

$${\phi }_1=\frac{\pi }{6};{\phi }_2=\frac{\pi }{6}$$ $$R=\frac{a}{2}\cot {\phi }_1=\frac{0.2}{2}\cot \frac{\phi }{6}=0.1X\sqrt{3}=0.1732m$$

and

$$\Rightarrow B=6B_a=\frac{6{\mu }_{0}I}{4\pi R}\left(\sin \frac{\pi }{6}+\sin \frac{\pi }{6}\right)$$

Putting values, we have :

$$\Rightarrow B=\frac{6X{10}^{-7}X10X1}{0.1732}==3.462X{10}^{-5}T$$

We shall determine magnetic field to different straight segments of wires. Let us consider the out of plane orientation as positive. Now, for wire segment AC, the point P is at the end of straight wire of length 4 m and is at a perpendicular linear distance of 4 m. The magnetic field at P due to segment AC is perpendicular and out of the plane of drawing. The magnetic field due to segment AC is :

$$B_{AC}=\frac{\sqrt{2}{\mu }_{0}I}{8\pi L}=\frac{\sqrt{2}{\mu }_{0}I}{8\pi X4}=\frac{\sqrt{2}{\mu }_{0}I}{32\pi }$$

For the wire segment CD, the point P lies on the extended line passing through the wire. The magnetic field due to this segment, therefore, is zero.

$$B_{CD}=0$$

For the wire segment DE, the point P is at the end of straight wire of length 2 m and is at a perpendicular linear distance of 2 m. The magnetic field at P due to segment DE is perpendicular and into the plane of drawing. The magnetic field is :

$$B_{DE}=-\frac{\sqrt{2}{\mu }_{0}I}{8\pi L}=-\frac{\sqrt{2}{\mu }_{0}I}{8\pi X2}=-\frac{\sqrt{2}{\mu }_{0}I}{16\pi }$$

For the wire segment EF, the point P is at the end of straight wire of length 2 m and is at a perpendicular linear distance of 2 m. The magnetic field at P due to segment DE is perpendicular and into the plane of drawing. The magnetic field is :

$$B_{EF}=-\frac{\sqrt{2}{\mu }_{0}I}{16\pi }$$

For the wire segment FG, the point P lies on the extended line passing through the wire. The magnetic field due to this segment, therefore, is zero.

$$B_{FG}=0$$

For the wire segment GA, the point P is at the end of straight wire of length 4 m and is at a perpendicular linear distance of 4 m. The magnetic field at P due to segment GA is perpendicular and out of the plane of drawing. The magnetic field is :

$$B_{GA}=\frac{\sqrt{2}{\mu }_{0}I}{32\pi }$$

The net magnetic field at P is :

$$B=B_{AC}+B_{CD}+B_{DE}+B_{EF}+B_{FG}+B_{GA}$$ $$\Rightarrow B=\frac{\sqrt{2}{\mu }_{0}I}{32\pi }+0-\frac{\sqrt{2}{\mu }_{0}I}{16\pi }-\frac{\sqrt{2}{\mu }_{0}I}{16\pi }+0+\frac{\sqrt{2}{\mu }_{0}I}{32\pi }$$ $$\Rightarrow B=-\frac{\sqrt{2}{\mu }_{0}I}{16\pi }$$ $$\Rightarrow B=-\frac{\sqrt{2}X4\pi X{10}^{-7}X10\sqrt{2}}{16\pi }=-\frac{2X{10}^{-7}X10}{4}$$ $$\Rightarrow B=-5X{10}^{-7}T$$

The net magnetic field is into the plane of drawing.