# Video: Calculating the Electromotive Force Induced in a Wire Moving in a Magnetic Field

A wire of length 5.0 cm moves at 3.0 m/s perpendicularly to a magnetic field that has a flux density of 10 T. What is the magnitude of the electromotive force induced? [A] 1.5 V [B] 30 V [C] 4.5 V [D] 10 V [E] 0.15 V

06:23

### Video Transcript

A wire of length 5.0 centimeters moves at 3.0 meters per second perpendicularly to a magnetic field that has a flux density of 10 teslas. What is the magnitude of the electromotive force induced? (a) 1.5 volts, (b) 30 volts, (c) 4.5 volts, (d) 10 volts, (e) 0.15 volts.

In our problem, we are told that the length of the wire is 5.0 centimeters. The letter 𝐿 is used to represent the length of the wire. We also know from our problem that our wire is moving at a speed of 3.0 meters per second. The letter 𝑣 represents the speed or velocity of an object. The flux density of our magnetic field has been stated to be 10 teslas. The letter 𝐵 is used to represent the flux density of a magnetic field. For this video, when we refer to the variable 𝐵, we’ll simply refer to it as the magnetic field.

We are solving for the magnitude of the induced electromotive force. The Greek letter 𝜀 is used to represent the electromotive force or as is commonly referred to the EMF. Now that we have made a list of our known variables and what we are solving for, we can draw a diagram of our problem. In our diagram, we have used yellow x’s to represent the magnetic field. Since the problem did not tell us the direction of the magnetic field, we chose to make the magnetic field into the screen and represented that with x’s. We have labeled the magnetic field as 𝐵 equals 10 teslas. We have drawn our wire as a vertical pink line on the left side of our diagram.

The problem does not state the orientation of the wire, only that the wire moves perpendicularly to the magnetic field. We have chosen to draw our wire vertically and have labeled it with the length 𝐿 is equal to 5.0 centimeters. The blue arrow in our diagram represents the velocity of the wire moving to the right perpendicularly to the magnetic field. The problem once again did not tell us the direction that the wire would be moving. We chose the direction to be to the right. And we labeled the magnitude as 𝑣 is equal to 3.0 meters per second.

To calculate the induced EMF for our problem, we must recall Faraday’s law. Faraday’s law states that the induced EMF, 𝜀, is equal to the number of turns in the wire, 𝑁, times the change in magnetic flux, ΔΦ, divided by the change in time, Δ𝑡. Although the problem is discussing a wire and not a closed circuit, an EMF can still be induced. This happens as the wire moves through the magnetic field as shown in the diagram and causes a change in magnetic flux. Because the wire does not have multiple loops to it, we can say that the 𝑁 or number of turns would be one. This leaves us with an expression for the induced EMF as the change in flux over the change in time.

Unfortunately, the problem did not give us the change in magnetic flux. To continue solving the problem, we must find an expression for the change in flux in terms of our known variables. We need to recall the equation for magnetic flux. The magnetic flux Φ is equal to the magnetic field 𝐵 times the area 𝐴 times the cos of the angle 𝜃. The area in the equation refers to the space within the field that is carved out by the moving wire. The angle 𝜃 refers to the angle between the area vector as carved out by our moving wire within the field and our magnetic field vector.

To determine an expression for our change in flux or ΔΦ, we must figure out which of our variables is changing. Looking at our magnetic field, it is a constant of 10 teslas, therefore, is a constant, and we can keep it as the variable 𝐵. Recall that the area is a space carved out by our moving wire within our field. At some later time 𝑡 after the wire has entered the magnetic field, it will have carved out an area as shown in the diagram. The length of the wire remains 𝐿 and the distance traveled into the field can be labeled as 𝑥. Notice that the pink shape in the picture that we have drawn is a rectangle.

Recall that the area of a rectangle is base times height. Based on the diagram, the area of this rectangle would be 𝑥 times 𝐿. The area vector is normal or perpendicular to the surface area of the carved-out space. Therefore, in this case, it would be into the screen or in the same direction as the magnetic field, making the angle between the area vector and the magnetic field vector zero degrees. Remember that the cos of zero degrees is one. Now that we have an expression for the change in flux, we can plug it back in to Faraday’s law.

The induced EMF is equal to the magnetic field times the distance times the length of the wire divided by the change in time. Recall from kinematics that the distance divided by the change in time is the speed. Replacing distance over change in time with speed, we get induced EMF is equal to the magnetic field times the speed of the wire times the length of the wire. We now have an expression for induced EMF that contains all of the known variables from our problem. Let’s go back to our known values and make sure that all of the units are in agreement with one another.

Looking at our length, we can see that it’s measured in centimeters but that our speed is measured in meters per second. This means that we must convert one of our units, but which one? Our final answers are measured in volts. A volt is defined as one kilogram meters squared per second cubed per ampere. Because our final answer contains meters squared hidden within the units, we need to convert our length from centimeters to meters to make sure that it is also in agreement. Because there are 100 centimeters in one meter, we can multiply our length by one meter over 100 centimeters. We see that we have both centimeters in the numerator and denominator, which means that they will cancel each other out.

In the numerator, we multiply out 5.0 times one and get 5.0. And in the denominator, we multiply one by 100 and we get 100. Because the centimeters canceled out, we are left with the units of meters. When you divide by 100, it’s the same thing as moving your decimal two places to the left. This gives us a length of 0.050 meters. Now that our units are in agreement, we can put all of our known values into our expression. For the magnetic field, we put in 10 teslas. For the speed, we put in 3.0 meters per second. And for the length, we put in 0.050 meters. When we multiply out our values, we get 1.5 volts.

So we can say that the magnitude of the electromotive force induced is 1.5 volts. Looking back at our answer choices, we can see that answer choice (a) says 1.5 volts.