Question Video: Comparing Induced emfs in an AC Generator | Nagwa Question Video: Comparing Induced emfs in an AC Generator | Nagwa

Question Video: Comparing Induced emfs in an AC Generator Physics • Third Year of Secondary School

In an AC generator, what is the ratio between the induced emf across a coil when the plane of the coil makes a 30° angle with the direction of the magnetic field and the maximum value of the induced emf?

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Video Transcript

In an AC generator, what is the ratio between the induced emf across a coil when the plane of the coil makes a 30-degree angle with the direction of the magnetic field and the maximum value of the induced emf? (A) One, (B) one-half, (C) square root of two over two, (D) square root of three over two.

In this question, we’re looking at an AC generator, which is essentially a coil of wire placed in a magnetic field. Let’s begin by drawing a diagram of the scenario.

We can imagine the magnetic field lines pointing to the right of our screen. And let’s say we have the coil of wire placed such that the plane of the coil is perpendicular to the direction of the magnetic field. This coil is in fact rectangular. But in the diagram, we are looking at our setup in such a way that we can only really see one side of the coil.

At this point, it’s worth recalling how to calculate the emf generated in a coil of wire rotating in a magnetic field. This is the equation we’re looking for. 𝜀 is equal to 𝑛𝐴𝐵𝜔 sin 𝜃, where 𝜀 is the electromotive force generated across our coil. 𝑛 is the number of turns of wire making up the coil. 𝐴 is its cross-sectional area. 𝐵 is the magnetic flux density or strength of the magnetic field. 𝜔 is the angular velocity or how fast the coil is rotating in the magnetic field. And 𝜃 is the angle between the magnetic field and a line that is normal to the plane of the coil.

In this question, we need to find the ratio between the emf induced when the plane of the coil is at 30 degrees to the direction of the magnetic field and the maximum possible emf induced in this coil. Let’s first notice that all of the quantities on the right-hand side of our emf equation are constants, apart from 𝜃. This is because the coil we are using does not change over time. So the number of turns and the cross-sectional area are constant. Similarly, the magnetic flux density is not changing, so this is constant too. And finally the coil in an AC generator will rotate at a constant angular velocity, so 𝜔 is constant too.

The only thing that is changing is the angle 𝜃. As the coil rotates, the angle between the magnetic field and the line normal to the plane of the coil constantly changes. As a result of this, the value of sin 𝜃 also changes. Hence, we can say that the ratio we are trying to find is equal to 𝑛𝐴𝐵𝜔 sin 𝜃 divided by 𝑛𝐴𝐵𝜔 sin 𝜃 max, where 𝜃 max is the angle that gives us the maximum possible induced emf.

Although we don’t yet know what this angle is, we can realize that the maximum possible emf value will be the value at which the sine of the angle is as large as possible. Now, the maximum possible value of sin 𝜃 is one, as the sine function varies between values of one and negative one. This happens when 𝜃 max is 90 degrees, in other words, when the plane of the coil is parallel to the magnetic field. Because remember 𝜃 is measured between the magnetic field direction and the normal to the plane of the coil. Hence, the denominator of our expression will simply become 𝑛𝐴𝐵𝜔.

Looking at the numerator then, we can find the value of the emf generated when the plane of the coil is at an angle of 30 degrees to the direction of the magnetic field. This does not simply mean 𝜃 is 30 degrees, because once again 𝜃 is measured from the normal to the plane of the coil and the question mentions that the plane of the coil itself is at 30 degrees to the field. In this situation, the angle 𝜃 will actually be equal to 90 degrees minus 30 degrees. So the numerator of our ratio becomes 𝑛𝐴𝐵𝜔 sin 60 degrees.

The value of the sin of 60 degrees is exactly square root of three divided by two, meaning our numerator simplifies to the square root of three 𝑛𝐴𝐵𝜔 divided by two. At this point, we can see that the factors of 𝑛𝐴𝐵𝜔 in the numerator and denominator cancel. And our ratio simplifies to the square root of three divided by two. This means we’ve arrived at our final answer.

The ratio between the induced emf across the coil when the plane of the coil makes a 30-degree angle with the direction of the magnetic field and the maximum value of the induced emf is square root of three divided by two. This corresponds to answer option (D).

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