Question

Two infinitely long conductors kept along XX' and YY' axes are carrying current \( I_1 \) and \( I_2 \) along -X axis and -Y axis respectively. Find the magnitude and direction of the net magnetic field produced at point P(X, Y).

Hint:

The magnetic fields from two perpendicular current-carrying wires add vectorially, and the total magnetic field at a point is given by the Pythagorean theorem when the fields are perpendicular.

Answer 1

The magnetic field at a point due to a current-carrying conductor is given by Ampere's law. The magnitude of the magnetic field \( B \) due to a current \( I \) in an infinitely long straight conductor at a distance \( r \) from the wire is: \[ B = \frac{\mu_0 I}{2 \pi r} \] Where: - \( \mu_0 \) is the permeability of free space, - \( I \) is the current, - \( r \) is the perpendicular distance from the wire to the point. 

Step 1: Magnetic Field due to \( I_1 \) (along the XX' axis) The magnetic field at point P due to the current \( I_1 \) will be circular around the conductor along the XX' axis. Since the current \( I_1 \) flows along the -X axis, the magnetic field at P will follow the right-hand rule. The direction of the magnetic field due to \( I_1 \) at point P is into the page. \[ B_1 = \frac{\mu_0 I_1}{2 \pi r_1} \] Where \( r_1 \) is the distance from the conductor along the XX' axis to point P. 

Step 2: Magnetic Field due to \( I_2 \) (along the YY' axis) The magnetic field at point P due to the current \( I_2 \) flowing along the -Y axis will be circular around the conductor along the YY' axis. Using the right-hand rule again, the direction of the magnetic field due to \( I_2 \) at point P is out of the page. \[ B_2 = \frac{\mu_0 I_2}{2 \pi r_2} \] Where \( r_2 \) is the distance from the conductor along the YY' axis to point P. 

Step 3: Net Magnetic Field at P The net magnetic field at point P is the vector sum of the magnetic fields \( B_1 \) and \( B_2 \). Since the directions of the magnetic fields due to \( I_1 \) and \( I_2 \) are perpendicular to each other (into and out of the page), the net magnetic field \( B_{\text{net}} \) can be found using the Pythagorean theorem: \[ B_{\text{net}} = \sqrt{B_1^2 + B_2^2} \] Substituting the expressions for \( B_1 \) and \( B_2 \): \[ B_{\text{net}} = \sqrt{\left(\frac{\mu_0 I_1}{2 \pi r_1}\right)^2 + \left(\frac{\mu_0 I_2}{2 \pi r_2}\right)^2} \] 

Step 4: Direction of the Net Magnetic Field The direction of the net magnetic field is determined by the vector sum of \( B_1 \) and \( B_2 \). Since \( B_1 \) is into the page and \( B_2 \) is out of the page, the net magnetic field will be in a direction perpendicular to both, forming an angle with respect to both axes. Thus, the magnitude and direction of the net magnetic field at point P is given by the above formula and can be determined from the geometry of the problem.