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, I1 = I2 = 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
\[B_{net} = \sqrt{{B_1}^2 + {B_2}^2 + 2 B_1 B_2 \cos\theta}\]
\[= \sqrt{\left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)^2 + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)^2 + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right) + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)\cos60^\circ}\]
\[= \sqrt{( {10}^{- 4} ) + ( {10}^{- 4} )^2 + 2( {10}^{- 4} )( {10}^{- 4} ) \times \frac{1}{2}}\]
\[ = \sqrt{3} \times {10}^{- 4} T\]
\[ = 1 . 732 \times {10}^{- 4} T\]
