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Two long, parallel wires are separated by a distance of 0.400 m (see the figure (Figure 1)). The currents I_1 and I_2 have the directions shown. Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the order. 6.00 times 10^-6 N Is the force attractive or repulsive?

User Cilap
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Answer:

The force is attractive because the currents in the two wires are in opposite directions.

Step-by-step explanation:

According to the given figure, the two wires are parallel to each other and are separated by a distance of 0.400 m.

The magnitude of the force (F) exerted by each wire on a 1.20-m length of the other wire can be calculated using the formula:

F = μ0 * I1 * I2 * L / (2 * π * d)

where μ0 is the permeability of free space, I1 and I2 are the currents in the wires, L is the length of the wire, and d is the distance between the wires.

Plugging in the given values, we get:

F = (4π * 10^-7 T*m/A) * (2.50 A) * (4.50 A) * (1.20 m) / (2 * π * 0.400 m)

F = 6.00 * 10^-6 N

The force is attractive because the currents in the two wires are in opposite directions.

User Aksu
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