0.3 Magnetic field due to current in straight wire  (Page 3/4)

Problem : A square loop of side “L” carries a current “I”. Determine the magnetic field at the center of loop.

Solution : The magnetic field due to each side of the square here is same as Magnetic field due to current in straight wire at a distance L/2 on the perpendicular bisector. The magnetic fields due to current in the four sides are in the same direction. Hence, magnitude of magnetic field due to current in loop is four times the magnetic field due to current in one side :

$$\Rightarrow B=4X\frac{{\mu }_{0}IL}{2\pi R\sqrt{\left(L^2+4R^2\right)}}$$

Here, R = L/2

Magnetic field at the center of square loop

$$\Rightarrow B=4X\frac{{\mu }_{0}IL}{2\pi \frac{L}{2}\sqrt{\{L^2+4{\left(\frac{L}{2}\right)}^2\}}}=\frac{4{\mu }_{0}IL}{\pi L\sqrt{2}L}$$ $$\Rightarrow B=\frac{2\sqrt{2}{\mu }_{0}I}{\pi L}$$

Problem : A current 10 ampere flows through the wire having configuration as shown in the figure. Determine magnetic field at P.

Magnetic field due to current in the arrangment

Solution : 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 :

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

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.

$$B_{\text{CD}}=0$$

For the wire segment DE, the angles between the line segment and line joining the point P with end points are known by geometry of the figure. Hence, Magnetic field due to this segment is :

$$B_{\text{DE}}=-\frac{{\mu }_{0}I}{4\pi R}\left(\cos {\theta }_1+\cos {\theta }_2\right)=-\frac{{\mu }_{0}I}{4\pi \sqrt{2}}X\left(\cos {45}^0+\cos {45}^0\right)$$ $$\Rightarrow B_{\text{DE}}=-\frac{{\mu }_{0}I}{4\pi \sqrt{2}}X\frac{2}{\sqrt{2}}=-\frac{{\mu }_{0}I}{4\pi }$$

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

$$B_{\text{EF}}=0$$

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 :

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

The net magnetic field at P is :

$$B=B_{\text{AC}}+B_{\text{CD}}+B_{\text{DE}}+B_{\text{EF}}+B_{\text{FA}}$$ $$\Rightarrow B=\frac{\sqrt{2}{\mu }_{0}I}{32\pi }+0-\frac{{\mu }_{0}I}{4\pi }+0+\frac{\sqrt{2}{\mu }_{0}I}{32\pi }$$ $$\Rightarrow B=\frac{{\mu }_{0}I\left(\sqrt{2}-4\right)}{16\pi }$$ $$\Rightarrow B=-\frac{2.59X4\pi X{10}^{-7}X10}{16\pi }=-\frac{2.59X{10}^{-7}X10}{4}$$ $$\Rightarrow B=-0.65X{10}^{-6}=-6.5X{10}^{-5}T$$

The net magnetic field is into the plane of drawing.