Video: Rate of Change of Induced Current

Video Transcript

A current is induced in a circular loop of radius 1.5 centimeters between two poles of a horseshoe electromagnet when the current in the electromagnet is varied. The magnetic field in the area of the loop is perpendicular to the loop and has a uniform magnitude. If the rate of change of the magnetic field is 10 teslas per second, find the magnitude of the induced current if the resistance of the loop is 25 ohms.

We’re told in this statement the rate of change of the magnetic field, 10 teslas per second, which we’ll call 𝑑𝐵 𝑑𝑡. We’re also told that the resistance of the circular loop is 25 ohms, what we’ll call capital 𝑅. And we’re told that the radius of this loop is 1.5 centimeters, what we’ll call lower case 𝑟. We want to solve for the magnitude of the current that’s induced in the loop. We can call that 𝐼.

Let’s begin by drawing a sketch of our loop in the magnet. We have a circular wire loop of radius lowercase 𝑟 and an electrical resistance of capital 𝑅, 25 ohms. Our loop is put between the poles of a horseshoe magnet, which is an electromagnet whose magnetism is created by running current through the loops that surround the magnet.

A magnetic field is created that passes through the circular loop. And the strength of this magnetic field changes in time. We’re told that the change in 𝐵 over 𝑡 is 10 teslas per second. To figure out the current that is induced in the circular loop due to this change in magnetic field, we can recall Faraday’s law. This law says that the emf, here symbolized with 𝜀, that’s induced in a wire is equal to the change in magnetic flux that passes through that closed loop per unit time multiplied by the number of loops in The wire.

And in support of this law we can further recall that magnetic flux, 𝜙 sub 𝑚, is equal to magnetic field strength, 𝐵, multiplied by the area through which that field as moving. When we apply Faraday’s law to our scenario, the first thing we notice is that 𝑁 is equal to one. We just have one single loop. The next element we want to look into is the magnetic flux, 𝜙 sub 𝑚. When we look at the relationship for that, we wanna see which of 𝐵 or 𝐴 or both are changing as this problem unfolds.

When we look at the circular wire loop through which the magnetic field lines pass, we see that the area of that loop doesn’t change throughout this problem. The magnetic field, 𝐵, however does change in time. And we’re told the rate of that change in the problem statement. This indicates that the change in the quantity 𝐵 times 𝐴 per change in time is equal to 𝐴 times 𝑑𝐵 𝑑𝑡 since 𝐴 is a constant value throughout.

We already know 𝑑𝐵 𝑑𝑡 because it’s given in the problem statement. And we can solve for 𝐴 based on knowing the radius of our circular loop, lower case 𝑟. Recalling that the area of a circle is equal to 𝜋 times its radius squared, we can write that the induced emf, or voltage, in the circular wire loop is equal to 𝜋 times 0.015 meters squared times 10 teslas per second.

If we calculated this value, it would give us an answer in units of volts. But we want an answer of current that is induced in this loop rather than the potential difference. Recalling Ohm’s law shows us that potential difference, 𝑉, is equal to current times resistance, 𝑅.

So to solve for the current, we can divide the induced emf by the resistance 𝑅, which we’re told is 25 ohms. When we enter these values on our calculator, we see that, to two significant figures, 𝐼 is 2.8 times 10 to the negative fourth amps. That’s the current that is induced in the circular wire loop.