Video Transcript
Two long, straight, parallel conducting wires are separated by distance 𝑑 equals 15 centimeters, as shown in the diagram. The wires both carry 1.2-ampere currents in the same direction. What is the magnitude of the force that a length 𝐿 equals 2.5 meters of each wire exerts on the other? Use a value of four 𝜋 times 10 to the negative seven henries per meter for the magnetic permeability of the region between the wires.
Okay, so in our diagram of these two long, straight, parallel conducting wires, the distance marked as 𝑑 is 15 centimeters and the currents 𝐼 one and 𝐼 two are both the same, and they’re equal to 1.2 amperes. We’re also told to consider a stretch of wire of length 𝐿 equals 2.5 meters. Over this length, the wires exert some mutual force on one another. We can call that force 𝐹, and we want to solve for its magnitude. To begin doing that, let’s clear some space on screen.
And we can recall that, in general, the force per unit length between two current-carrying wires that are parallel to one another is equal to the magnetic permeability of the space between them — this is typically the magnetic permeability of free space, 𝜇 naught — multiplied by the current in the one conductor times the current in the other conductor all divided by two times 𝜋 times the perpendicular distance between the wires 𝑟. In our scenario, we’re told that the magnetic permeability of the material between the wires is four times 𝜋 times 10 to the negative seven henries per meter. This is indeed the defined value of the permeability of free space 𝜇 naught. In other words, we’re treating our two conducting wires as though they’re separated by vacuum.
When we apply this relationship to our scenario, we’ll replace the perpendicular distance 𝑟 with the given perpendicular distance 𝑑. We want to solve this equation then for the force 𝐹. We can begin doing this by multiplying both sides by 𝐿, canceling that factor on the left, which then gives us this expression for the force 𝐹. Substituting in, we have four 𝜋 times 10 to the negative seven henries per meter for 𝜇 naught, 1.2 amperes for both currents 𝐼 one and 𝐼 two, 2.5 meters for the length 𝐿, and 15 centimeters for the separation distance 𝑑.
Note that we have distance units of meters in our numerator, but distance units of centimeters in our denominator. Before we calculate 𝐹, we’ll want these to agree. Let’s convert our distance units of centimeters to meters. One centimeter equals 10 to the negative two or one one hundredth of a meter. So 15 centimeters equals 15 times 10 to the negative two meters, which is the same as 0.15 meters. When we calculate 𝐹, we get an answer of exactly 4.8 times 10 to the negative six newtons. This is the magnitude of the force either wire exerts on the other current-carrying wire over a length span of 2.5 meters.
