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

An alternating current generator contains 25 rectangular loops of conducting wire with side lengths of 45 cm and 35 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 uniform magnetic field at 22 revolutions per second. The peak potential difference across the terminals is 105 V. What is the strength of the magnetic field? Give your answer to two decimal places.

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### Video Transcript

An alternating current generator contains 25 rectangular loops of conducting wire with side lengths of 45 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 22 revolutions per second. The peak potential difference across the terminals is 105 volts. What is the strength of the magnetic field? Give your answer to two decimal places.

Let’s begin by recalling that rotating a conducting loop in a magnetic field induces electromotive force or emf in the loop. And in the case of an AC generator like we have here, the potential difference across the terminals of the loop is the provided emf. Let’s also 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 here there are 25 of them. So 𝑛 equals 25.

𝐴 gives the area of each single loop in the generator. So to find this, we’ll multiply the loop’s side lengths together. But remember, we should be working in base SI units here. So since there are 100 centimeters in a meter, we’ll convert by moving the decimal place of the centimeter value one, two places to the left. Now, applying this to our side length values, 45 centimeters and 35 centimeters can be written as 0.45 meters and 0.35 meters. And their product gives an area value of 0.1575 meters squared.

Next in the formula is 𝐵, and it represents the strength of the magnetic field which we want to solve for. So let’s move on and look at 𝜔, angular frequency, which should be written in radians per second. But the value we have is given in revolutions per second. So to convert, recall that a revolution is referring to one full turn around a circle which measures two 𝜋 radians. So let’s make this substitution in the numerator and we have 𝜔 equals 22 times two 𝜋 or 44𝜋 radians per second.

Now, the next variable in the formula is time, but notice that the problem statement didn’t give us any value of time to work with. But there is one value we were given that we haven’t addressed yet. And that’s the peak emf, 105 volts. Recall that to maximize this formula or find the peak emf, we can just set this whole sine term equal to one, because that’s its maximum value. So we have a formula for peak emf that relates all these values, and we just have to solve it for 𝐵.

So let’s copy the formula down here. And notice that we flipped it so that 𝐵 is on the left-hand side. Now to get 𝐵 by itself, we’ll divide both sides of the formula by 𝑛𝐴𝜔. And finally, since they’re all written in base SI units, let’s go ahead and substitute in the values for peak emf 𝑛, 𝐴, and 𝜔. And we get a value of 0.1929 tesla. Finally, rounding our answer to two decimal places, we found that the strength of the magnetic field is 0.19 tesla.

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