The magnetic field \( B \) at a perpendicular distance \( r \) from an infinitely long straight conductor carrying current \( I \) is calculated using: \[ B = \frac{\mu_0 I}{2 \pi r} \] where \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)).
The perpendicular distance from point P to conductor 1 (along the X-axis) is \( X \). The magnetic field \( B_1 \) at P due to conductor 1 is: \[ B_1 = \frac{\mu_0 I_1}{2 \pi X} \] By the right-hand rule, with current along the -X axis, \( B_1 \) at P is directed out of the page.
The perpendicular distance from point P to conductor 2 (along the Y-axis) is \( Y \). The magnetic field \( B_2 \) at P due to conductor 2 is: \[ B_2 = \frac{\mu_0 I_2}{2 \pi Y} \] By the right-hand rule, with current along the -Y axis, \( B_2 \) at P is directed into the page.
Since \( B_1 \) and \( B_2 \) are perpendicular, the net magnetic field \( B_{\text{net}} \) is the vector sum, calculated using the Pythagorean theorem: \[ B_{\text{net}} = \sqrt{B_1^2 + B_2^2} = \sqrt{\left( \frac{\mu_0 I_1}{2 \pi X} \right)^2 + \left( \frac{\mu_0 I_2}{2 \pi Y} \right)^2} \]
The direction of the net magnetic field relative to the X-axis is found using trigonometry. Let \( \theta \) be the angle from the X-axis: \[ \tan \theta = \frac{B_2}{B_1} = \frac{\frac{\mu_0 I_2}{2 \pi Y}}{\frac{\mu_0 I_1}{2 \pi X}} = \frac{I_2 X}{I_1 Y} \] Therefore, the angle is: \[ \theta = \tan^{-1} \left( \frac{I_2 X}{I_1 Y} \right) \]
The magnitude of the net magnetic field at point P is: \[ B_{\text{net}} = \sqrt{\left( \frac{\mu_0 I_1}{2 \pi X} \right)^2 + \left( \frac{\mu_0 I_2}{2 \pi Y} \right)^2} \] The direction of the net magnetic field is at an angle \( \theta \) from the X-axis, where: \[ \theta = \tan^{-1} \left( \frac{I_2 X}{I_1 Y} \right) \]