Video Transcript
A 0.5-meter-long section of wire carrying a current of 12 amperes is positioned at 90 degrees to a magnetic field. The mass of the wire is 15 grams. What must the strength of the magnetic field be in order to counteract the weight of the wire? Use a value of 9.8 meters per second squared for the acceleration due to gravity.
So, here’s the idea. We have this half-meter-long stretch of wire. And this wire carries a current, we can call that current 𝐼, of 12 amperes. So, we’ll say 𝐼 equals 12 amps. And 𝑙, the length of the wire, is 0.5 meters. We’re told that this segment of wire exists within an external magnetic field. We can call this field capital 𝐵. And furthermore, we’re told that the mass of the wire, what we’ll call lowercase 𝑚, is equal to 15 grams. And that’s 15 grams, not kilograms.
Knowing all this, we want to solve for the strength of the magnetic field, that we’ve called 𝐵, that’s needed in order to counteract the weight of the wire. So, the idea there is that the weight of the wire, which we can draw as a force vector pointing down, where that weight is equal to the mass of the wire times the acceleration due to gravity, needs to be equal to, counteracted by the magnetic force upward on the wire. We can label that force 𝐹 sub 𝐵.
Along with being told the mass of the wire, we’re told the acceleration due to gravity, 9.8 meters per second squared. We can now write a force balance equation, since we’re told that the magnetic force is equal in magnitude to the weight force on the wire. We can say that 𝐹 sub 𝐵 is equal to 𝑚 times 𝑔. That’s what it means for the magnetic force to counteract the weight of the wire.
The question then is what is that magnetic force 𝐹 sub 𝐵? When we have a situation like we do in this exercise, where the current flowing through a straight wire is perpendicular to an externally applied magnetic field, then in that case, we can write an equation for the magnetic force on the wire. It’s equal to the current running through the wire multiplied by its length multiplied by the strength of the magnetic field it’s in.
Knowing this relationship, we can now substitute 𝐼 times 𝑙 times 𝐵 in for 𝐹 sub 𝐵. And when we do that, just to agree with the notation we’ve already picked for the length of the wire, which we’ve written as a script 𝑙, we’ll write it as 𝐼 script 𝑙 𝐵 is equal to 𝑚 times 𝑔. It’s a slightly different symbol but it means the same thing, the length of the wire.
Remember that this equation means that we are balancing the weight of the wire against the magnetic force, by setting these two sides equal to one another. That’s what we’re assuming. That means that the magnetic field here, capital 𝐵, will be that field strength which is needed to counteract the weight of the wire.
So, let’s rearrange this equation to solve for 𝐵 by itself. And to do that, we’ll divide both sides by the current multiplied by the length of the wire. Doing that cancels out both of these terms on the left-hand side of our expression. And we find that the magnetic field strength we want to solve for is equal to the mass of the wire times the acceleration due to gravity divided by the current through the wire multiplied by its length.
We know all four of these values. So, let’s plug them into this expression now. We’re just about ready to solve for 𝐵 except for one thing. Remember that our mass isn’t in units of kilograms. It’s in units of grams. If we leave it that way, that will cause a problem with this calculation because every other value we have here is in base units. For everything to be consistent with our numbers then, we’ll want this mass to be in base units as well. That unit is the kilogram.
So, recall that one kilogram is equal to 1000 grams. That’s another way of saying that one one thousandth of a kilogram is equal to one gram. Now, if we multiply both sides of this equation by 15, then we have that 15 grams is equal to fifteen one thousandth of a kilogram. Or, in other words, 0.015 kilograms.
When we’ve substituted that value in for the mass of our wire, now we’re ready to calculate 𝐵. When we do, we find the result of 0.0245 teslas, where a tesla, recall, is the unit of magnetic field. This, then, is the magnetic field strength that’s needed to counteract the weight of this current-carrying wire.
