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.
