### Video Transcript

Suppose the parallel plate capacitor shown is accumulating charge at a rate of 0.035 coulombs per second. What is the magnitude of the resulting induced magnetic field at a distance of 15 centimeters from the capacitor?

We can label this magnitude of magnetic field capital ๐ต. And as we study our diagram, we see that a current, ๐ผ, is flowing into our parallel plate capacitor and then out the other side. And we wanna solve for the magnetic field thatโs created a distance of 15 centimeters away from the center of these places.

Since the distance at which we wanna solve for the magnetic field is outside of our plates, that is, greater than our platesโ radius, then the magnetic field at this point is given by the relationship ๐ naught times ๐ผ over two ๐๐, where ๐ is the distance from the center of the plates. The value ๐ naught is a constant called the permeability of free space. This constant has a value of 1.26 times 10 to the negative sixth tesla meters per ampere.

Looking then at our equation for the magnetic field magnitude ๐ต, we know ๐ naught, and ๐ is given to us as 15 centimeters. That means we just need to solve for the current, ๐ผ, and then weโll be able to solve for the magnetic field.

We can recall that an ampere, which is the unit of current, is equal to a coulomb of charge passing a given point every second. And in our problem statement, weโre told this rate of charge per second that enters the plates. Since a coulomb per second is also an ampere, that means weโve been told the current.

We can now plug in our values, 1.26 times 10 to the negative sixth tesla meters per amp for ๐ naught, 0.035 amperes for our current ๐ผ, and, converting to units of meters, 0.15 meters for ๐. To two significant figures, this equals 4.7 times 10 to the negative eighth tesla. Thatโs the magnetic field strength produced by this parallel plate capacitor the given distance away from the plates.