Question

Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is 2.0 cm away from each of these wires.

Answer 1

the megnatic field at appoint which is 2.0 cm away from each of this wires

Answer 2

Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Then, the magnetic field at a point which is 2.0 cm away from each of these wires is $1.732 \times 10^{-4} T$

Explanation:

Step 1:

Given data :

Current Magnitude, $I_{1}=I_{2}=10 A$

Separating the point from the wires, d = 2 cm

Therefore, the magnetic field is supplied by current in the wire  

In the figure, the dotted circle shows the magnetic field lines placed in a plane perpendicular to the paper plane due to the current carrying wire.

 is a triangle of equilateral nature.

Due to current in the wire angle between magnetic fields, θ = 60°

Step 2:

∴ Magnet field Required at P

$B_{n e t}=\sqrt{B_{1}^{2}+B_{2}^{2}+2 B_{1} B_{2} \cos \theta}$

$=\sqrt{\left(\frac{2 \times 10^{-7} \times 10}{2 \times 10^{-2}}\right)^{2}+\left(\frac{2 \times 10^{-7} \times 10}{2 \times 10^{-2}}\right)^{2}+2 \frac{2 \times 10^{-7} \times 10}{2 \times 10^{-2}} \times \frac{2 \times 10^{-7} \times 10}{2 \times 10^{-2}} \cos 60^{\circ}}$

$=\sqrt{\left(\frac{10^{-7} \times 10}{10^{-2}}\right)^{2}+\left(\frac{10^{-7} \times 10}{10^{-2}}\right)^{2}+2 \frac{10^{-7} \times 10}{10^{-2}} \times \frac{10^{-7} \times 10}{10^{-2}} \cos 60^{\circ}}$

$=\sqrt{\left(10^{-4}\right)^{2}+\left(10^{-4}\right)^{2}+2 \times 10^{-4} \times 10^{-4} \times \frac{1}{2}}$

$=\sqrt{10^{-8}+10^{-8}+10^{-4} \times 10^{-4}}$

$=\sqrt{10^{-8}+10^{-8}+10^{-8}}$

$=\sqrt{3 \times 10^{-8}}$

$=10^{-4} \sqrt{3}$

$=1.732 \times 10^{-4} T$