Question

Two long straight conductors are placed 100 cm apart. The first conductor carries a current of 10 A, and the second conductor carries a current of 5 A, both in the same direction. What force do the conductors exert on a 0.4 m length of each other?

A) 2.0 N
B) 1.0 N
C) 0.5 N
D) 4.0 N

Answer 1

The conductors carrying currents of 10 A and 5 A placed 100 cm apart will exert a magnetic force of 4.0 x 10^-6 N on a 0.4 m length of each other.

Step-by-step explanation:

The question asks for the magnetic force experienced by two long straight conductors carrying currents of 10 A and 5 A respectively, which are placed 100 cm apart from each other. The length of interest on these conductors is 0.4 m. To calculate the force, we use the formula for the force between two parallel currents:

F = (µ0 / 2π) * (I1 * I2 / d) * L

where µ0 is the magnetic constant (4π x 10-7 T·m/A), I1 and I2 are the currents, d is the separation between the wires (100 cm or 1m), and L is the length of wire considered (0.4 m).

Based on this, we calculate:

F = (4π x 10-7 T·m/A / 2π) * (10 A * 5 A / 1 m) * 0.4 m = (2 x 10-7 N/A2) * 50 A2 * 0.4 m = 4.0 x 10-6 N

Since all the currents are flowing in the same direction and F is tiny, the conductors will exert an attractive force on each other.