Video Transcript
An alternating current generator contains 50 rectangular loops of conducting wire with side lengths of 55 centimeters and 35 centimeters, the ends of which form terminals. The sides of the loops with the same length as each other are parallel to each other. The loops rotate within a uniform magnetic field at 18 revolutions per second within a 360-millitesla uniform magnetic field. What is the root-mean-square voltage across the terminals? Give your answer to the nearest volt.
To begin, recall that moving a conducting loop in a magnetic field induces electromotive force or emf in the loop. In the case of a generator like we have here, the voltage across the wire terminals is the provided emf. To find the root mean square emf, we’ll use this formula: one over the square root of two times the peak emf. There are different ways to quantify emf because its value isn’t constant. Rather, it’s oscillating or moving back and forth within a range of values as time goes on and the wire loops spin. So we’ll need to find its maximum or peak value in order to find the root mean square.
Now, let’s clear some room and recall the formula for determining emf as a function of time: 𝑛 times 𝐴 times 𝐵 times 𝜔 times the sin of 𝜔 times 𝑡. 𝑛 represents the number of rotating loops in the generator. And we know that there are 50 of them, so 𝑛 equals 50. Next, 𝐴 represents the area of one single loop. So multiplying this by 𝑛, the number of loops, gives the combined area of all the loops in the generator. We’ve been told the side lengths of the loops, so we’ll use their measurements to find an area. But before we multiply them together, keep in mind that all the values we’re working with should be expressed in base SI units. So we’ll need to convert centimeters into plain meters.
Recall that there are 100 centimeters in one meter. And to convert, we’ll move the decimal place of the centimeter value one, two places to the left. Now, applying this to our side length values, 55 centimeters and 35 centimeters can be written as 0.55 meters and 0.35 meters. And their product gives an area value of 0.1925 meters squared. Moving on, 𝐵 represents the strength of the magnetic field, which we know is 360 millitesla. But remember, we should be in base SI units. So let’s convert this into plain tesla. The prefix milli- means we’re talking about thousands. So there are 1000 millitesla in one tesla. And to convert, we’ll move the decimal place of the millitesla value one, two, three places to the left. And applying this up here, we know the magnetic field has a strength of 0.360 tesla.
Next, that’s not W. It’s the lowercase Greek letter 𝜔. And it represents angular frequency. Recall that we measure angles with radians and frequency with per second. And per second is going to be the only SI unit associated with 𝜔. Although there are several ways to measure angles, there is no SI unit of angle. In general, it’s best to express angles using radians, but just remember that it isn’t an actual physical unit. Now, we’ve been told that the loops rotate at a rate of 18 revolutions per second. So let’s convert revolutions into radians.
Recall that one revolution measures one full turn around a circle or two 𝜋 radians. So let’s make this substitution in the numerator and we have 𝜔 equals 18 times two 𝜋 or 36𝜋 radians per second. And finally, although time appears as a variable in the formula, in this question we’re not going to need to know or use any specific value of time. Here’s why. In the formula, 𝑡 only appears within the sin function. And it’s the oscillating nature of sin as a function of time that’s causing this whole formula and therefore emf to have an oscillating value. And right now, because we’re trying to maximize or find the peak value of emf, we can just set the sin function to be equal to its maximum value, which is one.
So with the whole sin term set equal to one, we now have a formula for finding the peak emf. And we have a value for every term in the formula expressed in base SI units. So substituting in the values for 𝑛, 𝐴, 𝐵, and 𝜔, we get a peak emf value of about 391.9 volts. But remember, this isn’t our final answer. We calculated the peak emf so that we could find the root mean square.
Let’s bring back this formula, and we have that the root-mean-square emf equals one over the square root of two times the peak emf, which comes out to about 277.1 volts. And finally, rounding to the nearest volt, we have found that the root-mean-square voltage across the terminals is 277 volts.
