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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.
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The sides of the loops with the same length as each other are parallel to each other.
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The loops rotate within a uniform magnetic field at 22 revolutions per second.
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The peak potential difference across the terminals is 105 volts.
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What is the strength of the magnetic field?
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Give your answer to two decimal places.
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Let’s begin by recalling that rotating a conducting loop in a magnetic field induces electromotive force or emf in the loop.
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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.
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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.
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𝑛 represents the number of rotating loops in the generator.
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And we know here there are 25 of them.
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So 𝑛 equals 25.
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𝐴 gives the area of each single loop in the generator.
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So to find this, we’ll multiply the loop’s side lengths together.
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But remember, we should be working in base SI units here.
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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.
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Now, applying this to our side length values, 45 centimeters and 35 centimeters can be written as 0.45 meters and 0.35 meters.
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And their product gives an area value of 0.1575 meters squared.
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Next in the formula is 𝐵, and it represents the strength of the magnetic field which we want to solve for.
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So let’s move on and look at 𝜔, angular frequency, which should be written in radians per second.
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But the value we have is given in revolutions per second.
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So to convert, recall that a revolution is referring to one full turn around a circle which measures two 𝜋 radians.
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So let’s make this substitution in the numerator and we have 𝜔 equals 22 times two 𝜋 or 44𝜋 radians per second.
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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.
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But there is one value we were given that we haven’t addressed yet.
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And that’s the peak emf, 105 volts.
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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.
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So we have a formula for peak emf that relates all these values, and we just have to solve it for 𝐵.
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So let’s copy the formula down here.
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And notice that we flipped it so that 𝐵 is on the left-hand side.
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Now to get 𝐵 by itself, we’ll divide both sides of the formula by 𝑛𝐴𝜔.
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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.
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And we get a value of 0.1929 tesla.
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Finally, rounding our answer to two decimal places, we found that the strength of the magnetic field is 0.19 tesla.