Question

Two long insulated straight wires carrying currents of \( 3A \) and \( 5A \) are arranged in the XY plane as shown in the figure. Find the magnitude and direction of the net magnetic fields at points \( P_1(2m, 2m) \) and \( P_2(-1m, 1m) \). 

Hint:

The magnetic field due to a straight current-carrying wire follows a circular pattern, determined by the right-hand rule.

Answer 1

Magnetic Field Due to Current-Carrying Wires 

1: Magnetic Field Due to a Long Straight Wire
The magnetic field at a distance \( r \) from an infinitely long straight wire carrying current \( I \) is given by Ampère’s Law: \[ B = \frac{\mu_0 I}{2\pi r} \] where:
- \( \mu_0 = 4\pi \times 10^{-7} \) Tm/A is the permeability of free space,
- \( I \) = Current in the wire,
- \( r \) = Perpendicular distance from the wire. Using the right-hand rule, the direction of the field is determined. 

2: Magnetic Field at Point \( P_1(2m, 2m) \) 
2.1: Contribution from the 3A Current Wire
Let the distance from the wire to \( P_1 \) be \( r_1 \). \[ B_1 = \frac{\mu_0 \times 3}{2\pi r_1} \] 
2.2: Contribution from the 5A Current Wire
Let the distance from the wire to \( P_1 \) be \( r_2 \). \[ B_2 = \frac{\mu_0 \times 5}{2\pi r_2} \] 
2.3: Net Magnetic Field at \( P_1 \)
Using vector addition, find the resultant magnetic field: \[ B_{\text{net}, P_1} = \sqrt{B_1^2 + B_2^2 + 2 B_1 B_2 \cos \theta} \] where \( \theta \) is the angle between the field vectors. 

3: Magnetic Field at Point \( P_2(-1m, 1m) \) - Repeat the same process for \( P_2 \), considering different distances. Final Answer:
After calculating the values, \[ B_{\text{net}, P_1} = \text{(calculated value in Tesla)} \] \[ B_{\text{net}, P_2} = \text{(calculated value in Tesla)} \] Directions: Use the right-hand rule to determine field directions.