Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the same direction.What is the magnitude of the magnetic field halfway between the wires?

Question

Quizplus · Accepted Answer

The magnetic field due to a long straight wire carrying current is given by:B = (μ0 * I) / (2πr)where:μ0 is the permeability of free space (4π x 10^-7 T m/A)I is the current in the wire (in amperes)r is the distance from the wire (in meters)In this problem, we have two wires carrying currents of 10 A and 20 A, respectively. The wires are 40 cm apart, and we want to find the magnetic field halfway between them. The distance from each wire to the midpoint is 20 cm = 0.2 m.The magnetic field due to the first wire is:B1 = (μ0 * I1) / (2πr1)= (4π x 10^-7 T m/A) * (10 A) / (2π * 0.2 m)= 1.0 x 10^-5 TThe magnetic field due to the second wire is:B2 = (μ0 * I2) / (2πr2)= (4π x 10^-7 T m/A) * (20 A) / (2π * 0.2 m)= 2.0 x 10^-5 TThe total magnetic field at the midpoint is the sum of the magnetic fields due to each wire:B = B1 + B2= 1.0 x 10^-5 T + 2.0 x 10^-5 T= 3.0 x 10^-5 TTherefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10^-5 T.

Two Long Parallel Wires 40 Cm Apart Are Carrying Currents