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.
Figure

Answer 1

The magnitude of the magnetic field B at the points A1, A2, A3 are $0.67 \times 10^{-4} T$ , $2.67 \times 10^{-4} T$ , $2 \times 10^{-4} T$ and $1 \times 10^{-4} T$

Explanation:

To find the magnitude of the magnetic field B at the points A1, A2, A3:

In case of  point A1,

Current magnitude in wires, I = 10 A

Separating point A1 from the left side of the wire, d = 2 cm

Point A1 separation from the wire on the right side, d‘ = 6 cm

In the figure  attached below,

Red and blue arrow denotes magnetic field direction due to wire marked as red and blue respectively.

P (marked red) refers to the wire that carries current in a plane that goes into the paper.  

Q (marked blue) denotes the wire that carries current in a paper-based plane.  

We can see that too from the figure

$P A_{4}=Q A_{4}$

$\angle A_{4} A_{3} P=\angle A_{4} A_{3} Q=90^{\circ}$

$\angle A_{4} P A_{3}=\angle A_{4} Q A_{3}=45^{\circ}$

$\angle P A_{4} A_{3}=\angle Q A_{4} A_{3}=45^{\circ}$

$\angle P A_{4} Q=90^{\circ}$

Due to current in wires, the magnetic field at $A_1$ is given by

$B=\frac{\mu_{0} i}{2 \pi d}-\frac{\mu_{0} i}{2 \pi d^{\prime}} \ldots(1)$

$B=\frac{2 \times 10^{-7} \times 10}{2 \times 10^{-2}}-\frac{2 \times 10^{-7} \times 10}{6 \times 10^{-2}}$      

$B=\frac{10^{-7} \times 10}{10^{-2}}-\frac{10^{-7} \times 10}{3 \times 10^{-2}}$  

$B=\frac{10^{-7} \times 10}{10^{-2}}-\frac{10^{-7} \times 10}{3 \times 10^{-2}}$      

$B=10^{-5} \times 10-\frac{10^{-5} \times 10}{3}$      

$B=10^{-4}-\frac{10^{-4}}{3}$    

 $=\left(1-\frac{1}{3}\right) \times 10^{-4}$

$=\left(\frac{2}{3}\right) \times 10^{-4}$      

$=0.67 \times 10^{-4} T$

Likewise we use eq. (1) to get the magnetic field at $A_2.$

$B=\frac{2 \times 10^{-7} \times 10}{1 \times 10^{-2}}+\frac{2 \times 10^{-7} \times 10}{3 \times 10^{-2}}$    

$B=\frac{2 \times 10^{-7} \times 10}{10^{-2}}+\frac{2 \times 10^{-7} \times 10}{3 \times 10^{-2}}$      

$B=2 \times 10^{-4}+\frac{2 \times 10^{-4}}{3}$

$B=\left(2+\frac{2}{3}\right) 10^{-4}$

$B=\left(2+\frac{2}{3}\right) 10^{-4}$

$B=\left(\frac{6+2}{3}\right) 10^{-4}$

   $=\frac{8}{3} \times 10^{-4} T$

   $=2.67 \times 10^{-4} T$

Magnetic field in case of  $A_3$:

$B=\frac{2 \times 10^{-7} \times 10}{2 \times 10^{-2}}+\frac{2 \times 10^{-7} \times 10}{2 \times 10^{-2}}$

$B=\frac{10^{-7} \times 10}{10^{-2}}+\frac{10^{-7} \times 10}{10^{-2}}$

   $=2 \times 10^{-4} T$

Magnetic field in case of  $A_4$:

Divide point $A_4$ from wire on the left side, $d=\sqrt{2^{2}+2^{2}}=2 \sqrt{2} \mathrm{cm}$

Divide point $A_4$ from wire on the right side, $d^{\prime}=\sqrt{2^{2}+2^{2}}=2 \sqrt{2} \mathrm{cm}$

In this way, the magnetic field at $A_4$ is supplied by the current in the wires

$B=\sqrt{\left(\frac{2 \times 10^{-7} \times 10}{2 \sqrt{2} \times 10^{-2}}\right)^{2}+\left(\frac{2 \times 10^{-7} \times 10}{2 \sqrt{2} \times 10^{-2}}\right)^{2}}$      

$B=\sqrt{2\left(\frac{2 \times 10^{-7} \times 10}{2 \sqrt{2} \times 10^{-2}}\right)^{2}}$      

$B=\sqrt{2} \times \frac{2 \times 10^{-7} \times 10}{2 \sqrt{2} \times 10^{-2}}$      

$B=\frac{10^{-7} \times 10}{10^{-2}}$        

$B=10^{-5} \times 10$        

   $=1 \times 10^{-4} T$