Question

Figure shows two parallel wires separated by a distance of 4.0 cm and carrying equal currents of 10 A along opposite directions. Find the magnitude of the magnetic field B at the points A₁, A₂, A₃.

Answer 1

For point A₁,
Magnitude of current in wires, I = 10 A
Separation of point A₁ from the wire on the left side, d = 2 cm
Separation of point A₁ from the wire on the right side, d' = 6 cm



In the figure
Red and blue arrow denotes the direction of magnetic field due to the wire marked as red and blue respectively.
P (marked red) denotes the wire carrying current in a plane going into the paper.
Q (marked blue) denotes the wire carrying current in a plane coming out of the paper.
Also from the figure, we can see that
$$\begin{gathered} P{A}_4=Q{A}_4 \\ \angle A_4{A}_{3}P=\angle A_4{A}_{3}Q=90° \\ \Rightarrow \angle A_{4}P{A}_3=\angle A_{4}Q{A}_3=45° \\ \Rightarrow \angle P{A}_4{A}_3=\angle Q{A}_4{A}_3=45° \\ \Rightarrow \angle P{A}_{4}Q=90° \end{gathered}$$
The magnetic field at A₁ due to current in the wires is given by
$B=\frac{{\mu }_{0}I}{2\mathrm{\pi }d}-\frac{{\mu }_{0}I}{2\mathrm{\pi }d\text{'}}$ ...(1)

$$\begin{gathered} \Rightarrow B=\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}-\frac{2\times {10}^{-7}\times 10}{6\times {10}^{-2}} \\ =\left(1-\frac{1}{3}\right)\times {10}^{-4} \\ =0.67\times {10}^{-4}\mathrm{T} \end{gathered}$$

Similarly, we get the magnetic field at A₂ using eq. (1).
$$\begin{gathered} B=\frac{2\times {10}^{-7}\times 10}{1\times {10}^{-2}}+\frac{2\times {10}^{-7}\times 10}{3\times {10}^{-2}} \\ =\frac{8}{3}\times {10}^{-4}\mathrm{T} \\ =2.67\times {10}^{-4}\mathrm{T} \end{gathered}$$

Now,
Magnetic field at A₃:
$$\begin{gathered} B=\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}+\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}} \\ =2\times {10}^{-4}\mathrm{T} \end{gathered}$$

Magnetic field at A₄:
Separation of point A₄ from the wire on the left side, d = $\sqrt{2^2+2^2}=2\sqrt{2}\mathrm{cm}$
Separation of point A₄ from the wire on the right side, d' =$\sqrt{2^2+2^2}=2\sqrt{2}\mathrm{cm}$
Thus, the magnetic field at A₄ due to current in the wires is given by
$$\begin{gathered} B=\sqrt{{\left(\frac{2\times {10}^{-7}\times 10}{2\sqrt{2}\times {10}^{-2}}\right)}^2+{\left(\frac{2\times {10}^{-7}\times 10}{2\sqrt{2}\times {10}^{-2}}\right)}^2} \\ =1\times {10}^{-4}\mathrm{T} \end{gathered}$$