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.

#### Solution

Given:

Magnitude of currents, I_{1}_{ }= I_{2} = 10 A

Separation of the point from the wires, d = 2 cm

Thus, the magnetic field due to current in the wire is given by

\[B_1 = B_2 = \frac{\mu_0 I}{2\pi d}\]

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

From the figure, we can see that \[∆ P I_1 I_2\] is an equilateral triangle.

\[\angle I_1 P I_2 = {60}^\circ \]

Angle between the magnetic fields due to current in the wire, θ = 60°

∴ Required magnetic field at P

\[B_{net} = \sqrt{{B_1}^2 + {B_2}^2 + 2 B_1 B_2 \cos\theta}\]

\[= \sqrt{\left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)^2 + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)^2 + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right) + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)\cos60^\circ}\]

\[= \sqrt{( {10}^{- 4} ) + ( {10}^{- 4} )^2 + 2( {10}^{- 4} )( {10}^{- 4} ) \times \frac{1}{2}}\]

\[ = \sqrt{3} \times {10}^{- 4} T\]

\[ = 1 . 732 \times {10}^{- 4} T\]