# Question Video: Alternating Current Circuits Physics • 9th Grade

An alternating current generator contains 10 rectangular loops of conducting wire with side lengths of 25 cm and 23 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 within a 560 mT uniform magnetic field. The peak potential difference across the terminals is 85 V. How many revolutions per second do the loops turn through? Give your answer to the nearest revolutions per second.

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### Video Transcript

An alternating current generator contains 10 rectangular loops of conducting wire with side lengths of 25 centimeters and 23 centimeters, 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 within a 560-millitesla uniform magnetic field. The peak potential difference across the terminals is 85 volts. How many revolutions per second do the loops turn through? Give your answer to the nearest revolutions per second.

Let’s begin by recalling that rotating a conducting loop in a magnetic field induces electromotive force, or emf, in the loop equal to the potential difference across the loop’s terminals. Also, recall that as time goes on and the wire loops rotate, emf varies sinusoidally among a range of values. We can determine the maximum or peak emf using the formula 𝑛 times 𝐴 times 𝐵 times 𝜔, where 𝑛 is the number of rotating loops in the generator, 𝐴 is the area of each loop, 𝐵 is the strength of the magnetic field, and 𝜔 represents angular frequency, which we’re going to solve for to find how many revolutions per second the loops turn through.

Also, we’ve been told that the peak emf is 85 volts. So let’s go ahead and add that to our list of known values over here. Now, working our way up, we know that the magnetic field has a strength of 560 milliteslas, but all of our values should be expressed in base SI units. So let’s convert milliteslas into plain teslas. Now, since there are 1000 milliteslas in one tesla, we can convert by moving the decimal place of the millitesla value three places to the left. And applying this to our 𝐵-value, we can tell that 560 milliteslas equals 0.560 teslas.

Next, we’ll calculate the area of a rectangular loop using its side lengths, but those values are currently written in centimeters. So recall that we can convert centimeters into plain meters by moving the decimal place of the centimeter value two places to the left. Therefore, 23 centimeters and 25 centimeters can be written as 0.23 meters and 0.25 meters. Now, multiplying them together, we get an area value of 0.0575 meters squared. And finally, 𝑛 represents the number of loops in the generator. And we know there are 10. So 𝑛 equals 10.

Now that we have a value for every variable in the formula except angular frequency, let’s copy the formula over here and solve it for 𝜔. To do this, we’ll divide both sides of the formula by 𝑛𝐴𝐵. And finally, we’re ready to find 𝜔 by substituting in the values for peak emf, 𝑛, 𝐴, and 𝐵. Now, calculating, we have 𝜔 equals 263.98 radians per second. Now, angular frequency is typically expressed in radians per second. But in this question, we wanna give our answer in revolutions per second.

To convert, recall that one revolution refers to one full turn around a circle, which measures two 𝜋 radians. So let’s multiply 𝜔 by this conversion factor, which itself is just equal to one. And now we can cancel radians out of the numerator and denominator. So, now, once we divide this value by two 𝜋, we have 42.01 revolutions per second. And finally, rounding our answer to the nearest whole number, we have found that the loops in the AC generator turned through 42 revolutions per second.