Two very long, straight, and insulated wires are kept at 90° angle from each other in xy-plane as shown in figure. These wires carry currents of equal magnitude I, whose direction are shown in the figure. The net magnetic field at point P will be:

Based on the principle of superposition and the Right-Hand Thumb Rule, the net magnetic field at point P is calculated as follows:

The magnetic field $\vec{B}$ due to a long straight wire at a distance $d$ is given by:

$$B = \frac{\mu_0 i}{2\pi d}$$

Assuming point P is at an equal distance $d$ from both wires in the xy-plane:

• For Wire 1 (along the y-axis): Using the Right-Hand Thumb Rule, the magnetic field at point P acts out of the page (along the +z direction).$$\vec{B}_1 = \frac{\mu_0 i}{2\pi d} \hat{k}$$
• For Wire 2 (along the x-axis): The magnetic field at point P acts into the page (along the -z direction).$$\vec{B}_2 = -\frac{\mu_0 i}{2\pi d} \hat{k}$$

Net Magnetic Field:

The total magnetic field is the vector sum of the individual fields:

$$\vec{B}_{\text{net}} = \vec{B}_1 + \vec{B}_2$$

$$\vec{B}_{\text{net}} = \frac{\mu_0 i}{2\pi d} \hat{k} - \frac{\mu_0 i}{2\pi d} \hat{k}$$

$$\boxed{\vec{B}_{\text{net}} = 0}$$