### Video Transcript

A loop of conducting wire has a radius of 28 centimeters. The loop is in a uniform magnetic field of strength 125 milliteslas that is out of the plane of the diagram shown and parallel to the axis of the loop. The loop is rotated in 0.45 seconds to face in a direction that is at an angle of 65 degrees from its original axial direction. What is the magnitude of the electromotive force induced in the loop? Give your answer to two decimal places.

In our diagram, we see the uniform magnetic field pointing out of the screen. And we also see the initial position of our conducting loop, shown here in the dashed line, and its final position after it’s been rotated 65 degrees from its original orientation. Initially, the full cross-sectional area of this circular loop is exposed to the uniform magnetic field. As the loop rotates though, the magnetic flux through it changes. If the symbol Φ sub 𝐵 represents magnetic flux, then that flux is equal to the area 𝐴 exposed to a uniform magnetic field 𝐵 multiplied by that field strength.

We can see then that for a given conductor in a magnetic field, if either the magnetic field strength 𝐵 or the area of the conductor exposed to the field 𝐴 changes, then that will lead to a change in magnetic flux Φ sub 𝐵 experienced by the conductor. In our scenario, we have a constant magnetic field 𝐵, but the area 𝐴 of our loop exposed to that field does change. When the magnetic flux through a loop changes over time, an electromotive force is induced in the loop. This happens according to a law known as Faraday’s law.

Faraday’s law tells us that the emf induced in a conductor through which the magnetic flux changes is equal to negative the number of turns 𝑁 in the conductor multiplied by the change in magnetic flux ΔΦ sub 𝐵 through it all divided by the change in time Δ𝑡, over which that change in flux occurs. It’s this induced electromotive force, specifically the magnitude of that force, that we want to solve for. As a side note, emf is actually not a force, but rather is a potential difference expressed in units of volts.

To help us solve for emf induced in this loop, let’s record some of the information given to us. The radius of our conducting loop — we’ll call that radius 𝑟 — is 28 centimeters. The loop moves through a uniform magnetic field we’ll call 𝐵 of 125 milliteslas, and it goes through its rotation in a time Δ𝑡 of 0.45 seconds. That rotation is through an angle we’ll call 𝜃 of 65 degrees. Knowing all this, let’s clear some space on screen. And considering this equation we want to solve, we can note that 𝑁, the number of turns in our coil, is one. That is, the coil is just a single loop of wire. Therefore, we can leave 𝑁 out of this equation.

Recalling that magnetic flux Φ sub 𝐵 is equal to the magnetic field strength 𝐵 times the area exposed to that field 𝐴, we can replace Φ sub 𝐵 in our equation with 𝐵 times 𝐴. Our scenario shows us that the magnetic field strength 𝐵 is constant. However, the area of our loop exposed to that field does change over time. This fact that 𝐵 is constant while the area exposed to the magnetic field does change means that we can rewrite our expression for 𝜀. It’s equal to negative 𝐵 times Δ𝐴 divided by Δ𝑡.

The next question is, what is Δ𝐴? What is that change in the area of our loop exposed to the magnetic field? Imagine we’re looking at our magnetic field from a side-on view. From this perspective, the initial position of our conducting loop would look like this. Let’s call the area of this loop exposed to the magnetic field 𝐴 sub 𝐼. Since the loop is completely perpendicular to the field, that means its full area is available for magnetic field lines to pass through. Since in general, the area of a circle equals 𝜋 times the circle’s radius squared. We can write that 𝐴 sub 𝐼 the initial area of our loop exposed to the magnetic field is 𝜋 times 𝑟 squared, where 𝑟 is 28 centimeters.

Our loop of wire doesn’t stay that way we know, but rather it rotates through an angle we’ve called 𝜃. Note that arranged this way, fewer magnetic field lines are capable of passing through our circular loop. We’ll call this exposed area of our loop 𝐴 sub 𝑓. It’s equal to the initial area of our loop, that’s 𝜋 times 𝑟 squared, multiplied by the cos of the rotation angle 𝜃.

To convince ourselves that the cosine is the correct trigonometric function to use here, let’s note that if 𝜃 were 90 degrees, that is, if our loop of conducting wire were arranged like this, then no flux lines could pass through it. And indeed, the cos of 90 degrees is zero. Similarly, if 𝜃 were zero degrees, if our loop weren’t rotated at all, then the area of the loop exposed to the magnetic field would revert to 𝐴 sub 𝐼. The cos of zero degrees is one.

We can now write an expression for Δ𝐴, the change in the area of our loop exposed to the magnetic field. It’s equal to the final area 𝐴 sub 𝑓 minus the initial area 𝐴 sub 𝐼, or in other words 𝐴 sub 𝐼 times the cos of 𝜃 all minus 𝐴 sub 𝐼. We can then factor out 𝐴 sub 𝐼 from both of these terms as well as recognize that 𝐴 sub 𝐼 is equal to 𝜋 times 𝑟 squared. Now let’s take this whole expression and substitute it in for Δ𝐴 in our equation. In the resulting expression, notice that if we multiply through by this negative sign, that will effectively switch the order of the terms in parentheses here. That is, without the negative sign in front, in parentheses we have one minus the cos of 𝜃.

Note that all four of the variables in this expression are variables whose values we know. The magnetic field strength 𝐵 is 125 milliteslas, the radius of our loop 𝑟 is 28 centimeters, the angle 𝜃 is 65 degrees, and the time Δ𝑡 is 0.45 seconds. Before we calculate this expression, we’ll want to convert some units, milliteslas to teslas and centimeters to meters.

We can recall that the prefix milli- indicates 10 to the negative three or one one thousandth of a unit. And so, 125 milliteslas is equal to 125 times 10 to the negative three teslas or simply 0.125 teslas. In a similar way, the prefix centi- indicates one one hundredth or 10 to the negative two of a quantity, indicating that 28 centimeters is 0.28 meters. We’re now ready to calculate the induced emf 𝜀. Rounding our answer to two decimal places, we get a result of 0.04 volts. This is the magnitude of the emf induced in our rotating loop of wire.