Question Video: Alternating Current Circuits Physics

Video Transcript

An ac generator has a frequency of 50 hertz. The coil of the generator is initially parallel to the generator’s magnetic field, and its loops are coplanar. At what time after the coil starts rotating is the instantaneous potential difference across the coil equal to the root mean square potential difference produced by the generator?

Let’s begin by recalling that moving a conducting loop in a magnetic field induces electromotive force, or emf, in the loop. In an ac generator, like we have here, the potential difference across the loop’s terminals is the provided emf. Next, recall the formula for determining the instantaneous value of the alternating emf: 𝑛 times 𝐴 times 𝐵 times 𝜔 times the sin of 𝜔 times 𝑡, where 𝑛 is the number of rotating loops, 𝐴 is the area of each loop, 𝐵 is the strength of the magnetic field, 𝜔 is the angular frequency of the loops, and 𝑡 is time. Let’s also recall the formula for finding the root mean square emf: one over the square root of two times the peak or maximum emf.

Now notice the sin function in the instantaneous formula. It means as time goes on in the loop’s spin, emf varies sinusoidally. Let’s look at a diagram with a top-down view to visualize this. As viewed from above, here are the loops, which we know are coplanar or all facing the same direction. We also know that they’re initially parallel to the magnetic field. So, at time 𝑡 equals zero, emf is also zero since the magnetic field isn’t able to actually pass through the loops. This corresponds to setting 𝑡 equal to zero in the instantaneous emf formula. The sin of zero is zero, so there isn’t any induced emf in the wire loops at that first moment. Conversely, though, emf’s magnitude is maximized at the moments when the magnetic field can pass through the most area of the loop possible, which is when the plane of the loop is perpendicular to the field.

Looking back at the formula, we know that in order for emf to be maximized, the sin function has to be maximized. And the largest value sin can have is one. So, we can write a formula expressing that the maximum or peak emf happens when this entire term is equal to one. And this side of the formula just becomes 𝑛𝐴𝐵𝜔. Let’s make a substitution in the root mean square formula up here. And we now have full expressions for both instantaneous and root mean square emf. Now, we wanna determine the time at which instantaneous emf equals root mean square emf. So, we’ll set these two formulas equal to each other.

And now that we have our equality, let’s simplify by dividing both sides by 𝑛𝐴𝐵𝜔. Those terms cancel out. And now, we have the sin of 𝜔𝑡 equals one over the square root of two. And all we have to do is solve this for 𝑡 because it represents the time that satisfies the condition we’re looking for. Now, to undo the sin function that time appears in, we take the inverse sin of both sides of the formula. And remember, we should be measuring angles in radians. So, we write the inverse sin of one over the square root of two as 𝜋 over four radians. Now, to get 𝑡 by itself, we divide both sides of the formula by 𝜔. And we have 𝑡 equals 𝜋 over four radians divided by 𝜔, which we know represents the angular frequency of the spinning loops.

Now, we were told that the loops rotate at a rate of 50 hertz. So, recall that hertz is the SI unit of frequency, and it measures some number of occurrences per second. So, here, 50 hertz is referring to the loops making 50 full rotations or revolutions per second. Now, we’re gonna use this value to calculate. And the second is the base SI unit of time. So, the denominator is good to go. But remember that 𝜔 doesn’t just represent any frequency. It represents angular frequency. And we measure angles using radians, not revolutions. So, we need to make a conversion. Recall that one revolution means one full turn around a circle, which measures two 𝜋 radians. So, let’s make this substitution in the numerator, and we have 50 times two 𝜋 or 100𝜋 radians per second. Now, this is a proper angular frequency, and we’re all set to calculate.

So, 𝑡 equals 𝜋 over four radians divided by 100𝜋 radians per second or simplifying 𝜋 radians divided by 400𝜋 radians per second. Now, we can cancel 𝜋 radians from the numerator and denominator, leaving one over 400 per seconds. And units of per seconds in the denominator equals plane units of seconds in the numerator. So, our answer is one four hundredth of a second or 0.0025 seconds. And we did it. We found that the potential difference across the coil equals the root mean square potential difference produced by the generator 0.0025 seconds after the coil starts rotating.