Question Video: Alternating Current Circuits Physics • 9th Grade

An alternating current generator contains 5 rectangular loops of conducting wire with side lengths 15 cm and 25 cm, the ends of which form terminals. The sides of the loops with the same lengths as each other are parallel to each other. The loops rotate at 15 revolutions per second within a 620 mT uniform magnetic field. What is the peak potential difference across the terminals? Give your answer to two decimal places.

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Video Transcript

An alternating current generator contains five rectangular loops of conducting wire with side lengths 15 centimeters and 25 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 at 15 revolutions per second within a 620-millitesla uniform magnetic field. What is the peak potential difference across the terminals? Give your answer to two decimal places.

Here we have an AC generator which moves conducting loops in a magnetic field, which induces electromotive force, or emf, in the loop. And here the emf in the loop equals the potential difference across the loop’s terminals. Recall that we can calculate the instantaneous emf using the formula 𝑛 times 𝐴 times 𝐵 times 𝜔 times the sin of 𝜔 times 𝑡. And take special notice that as the wire loops spin and time goes on, emfs value varies sinusoidally. But in this question, we’re not necessarily concerned with emf as a function of time because our goal is to find the peak potential difference, or peak emf.

Looking at the instantaneous formula, these four terms which represent the number of loops, the area of each loop, the magnetic field strength, and the angular frequency of the loops all have constant unchanging values. It’s this sine term here that tells how emf varies as a function of time. So since we want to find the peak, or maximum, emf, we should set the sine term equal to its maximum value, which is just one. So the formula 𝑛 times 𝐴 times 𝐵 times 𝜔 will give us the answer we’re looking for. Now let’s get organized and make sure we have a value for each one of these terms expressed in base SI units so we can substitute them into the formula.

Now 𝑛 is a unitless number because it just represents the number of loops in the generator. Here, we know that’s five. So 𝑛 equals five. Next, 𝐴 is area, so we should have a value in square meters, and we’ll use length times width to find the area of a loop. But the side lengths were given to us in centimeters, so let’s convert them into meters. Since there are 100 centimeters in a meter, we can convert by moving the decimal point of the centimeter value one, two places to the left. So we’ll write 25 centimeters as 0.25 meters and 15 centimeters as 0.15 meters. And their product gives an area value of 0.0375 meters squared.

Next, 𝐵 measures the strength of the magnetic field, whose SI unit is the tesla. And right now we know its value in milliteslas, so recall that the prefix milli- means thousandth or 10 to the minus three. So if we move the decimal place of the millitesla value three places to the left, we know the strength of the magnetic field is 0.620 tesla. Moving on, 𝜔 represents angular frequency, which should be represented in radians per second. But we know the loops rotate at 15 revolutions per second. So let’s convert revolutions to radians. One revolution refers to one full turn around a circle, which measures two 𝜋 radians. So we can make this substitution in the numerator and we have 15 times two 𝜋, or 30𝜋, radians per second.

Now that we have all our values, let’s copy the formula down here and substitute in 𝑛, 𝐴, 𝐵, and 𝜔. And finally calculating, the peak emf equals 10.956 volts. So, rounding to two decimal places, we found that the peak potential difference across the terminals is 10.96 volts.

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