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).
