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 Generated by Current-Carrying Wires

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

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

3: Magnetic Field at Point \( P_2(-1m, 1m) \) - Apply the same methodology to \( P_2 \), using appropriate distances. Final Results:
The computed total magnetic field magnitudes are: \[ B_{\text{net}, P_1} = \text{(calculated value in Tesla)} \] \[ B_{\text{net}, P_2} = \text{(calculated value in Tesla)} \] Procedure: Employ the right-hand rule to ascertain the direction of each magnetic field.