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

Net Magnetic Field at Point P Due to Two Infinitely Long Conductors

Given:

• Conductor 1: Carries current \( I_1 \) along the -X axis (along XX' axis)
• Conductor 2: Carries current \( I_2 \) along the -Y axis (along YY' axis)
• Point P is at coordinates \( (X, Y) \)

Solution:

1. Magnetic Field Due to a Long Straight Conductor:

The magnetic field \( B \) at a distance \( r \) from a long straight conductor carrying current \( I \) is given by the formula: \[ B = \frac{\mu_0 I}{2 \pi r} \] where: - \( B \) is the magnetic field, - \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)), - \( I \) is the current in the conductor, - \( r \) is the perpendicular distance from the conductor to the point where the magnetic field is being calculated.

2. Magnetic Field Due to Conductor 1 (along XX' axis):

The distance of point P from conductor 1 (along the XX' axis) is \( X \) (since the conductor is along the \( X \)-axis). The magnetic field at point P due to conductor 1, using the formula, is: \[ B_1 = \frac{\mu_0 I_1}{2 \pi X} \] The direction of the magnetic field can be found using the right-hand rule. Since the current \( I_1 \) flows along the -X axis, the magnetic field at point P will be directed out of the page (towards the viewer) by the right-hand rule.

3. Magnetic Field Due to Conductor 2 (along YY' axis):

The distance of point P from conductor 2 (along the YY' axis) is \( Y \) (since the conductor is along the \( Y \)-axis). The magnetic field at point P due to conductor 2 is: \[ B_2 = \frac{\mu_0 I_2}{2 \pi Y} \] Again, applying the right-hand rule, since the current \( I_2 \) flows along the -Y axis, the magnetic field at point P will be directed into the page (away from the viewer).

4. Resultant Magnetic Field:

The magnetic fields \( B_1 \) and \( B_2 \) are perpendicular to each other (one is out of the page and the other is into the page), so we can calculate the net magnetic field using the Pythagorean theorem: \[ B_{\text{net}} = \sqrt{B_1^2 + B_2^2} \] Substituting the values of \( B_1 \) and \( B_2 \): \[ B_{\text{net}} = \sqrt{\left( \frac{\mu_0 I_1}{2 \pi X} \right)^2 + \left( \frac{\mu_0 I_2}{2 \pi Y} \right)^2} \]

5. Direction of the Net Magnetic Field:

The direction of the net magnetic field can be found using the right-hand rule applied to the two perpendicular fields. The resultant magnetic field will be directed at an angle \( \theta \) from the \( X \)-axis (where \( \theta \) is the angle between the net magnetic field and the \( X \)-axis). Using trigonometry: \[ \tan \theta = \frac{B_2}{B_1} = \frac{\frac{\mu_0 I_2}{2 \pi Y}}{\frac{\mu_0 I_1}{2 \pi X}} = \frac{I_2 X}{I_1 Y} \] Thus, the angle \( \theta \) is given by: \[ \theta = \tan^{-1} \left( \frac{I_2 X}{I_1 Y} \right) \]

Conclusion:

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: \[ \theta = \tan^{-1} \left( \frac{I_2 X}{I_1 Y} \right) \]