Video: Combined Effects of Magnetic and Electric Fields on Charged Particle Motion

An electron with a kinetic energy of 2.000 keV moves along a path that is equidistant from the surfaces of two parallel plates that are 1.00 cm apart and have a potential difference of 300 V. What is the strength of the uniform magnetic field acting parallel to the plates?

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Video Transcript

An electron with a kinetic energy of 2.000 kiloelectron volts moves along a path that is equidistant from the surfaces of two parallel plates that are 1.00 centimetres apart and have a potential difference of 300 volts. What is the strength of the uniform magnetic field acting parallel to the plates?

We can call the kinetic energy of the electron 2.000 kiloelectron volts 𝐾𝐸. The distance the parallel plates are apart 1.00 centimetres we’ll name 𝑑. And the potential difference between the plates 300 volts we’ll call 𝑉. We want to solve for the magnitude of the magnetic field that acts parallel to the plates. We’ll call this 𝐵.

We can start by drawing a diagram. We’re told we have two parallel plates separated by a distance 𝑑, through which an electron moves parallel to the plates’ surfaces. We’re also told that the plates have a potential difference 𝑉 across them. And if we assign the top plate to be positively charged and the bottom plate to be negatively charged, then we know that potential difference will create an electric field we’ve called 𝑒 that points from the top plate to the bottom plate.

The existence of the electric field tells us that this electron as it moves along parallel to the plates feels an electric force due to the field. It will be drawn upward toward the positive plate by an electric force. But because the electron’s path continues parallel to the plates, we know there must be another force at play that balances out the electric force. And in fact, there is. There is a magnetic force that acts on the electron pointing the opposite way.

These two forces, the electric and magnetic, balance one another out. So we can write the magnitude of 𝐹 sub 𝐸 is equal to the magnitude of 𝐹 sub 𝐵. At this point, we can recall the mathematical equations for electric and magnetic forces on a charged particle. The electric force 𝐹 sub 𝐸 is equal to the charge of a particle times the magnitude of the electric field it’s in. So we can write that the magnitude of the charge of the electron times the field that it’s in is equal to 𝐹 sub 𝐵.

When it comes to magnetic force, that’s equal to an object’s charge multiplied by its velocity times the magnitude of the magnetic field that’s in 𝐵. So the charge of an electron times the electric field magnitude is equal to the magnitude of the charge of the electron times its speed 𝑣 times the field 𝐵.

Since the electron’s charge appears on both sides of the equation, we can cancel it out and then drop the absolute value bars. So electric field 𝐸 equals charge speed 𝑣 times magnetic field 𝐵 or 𝐵 equals 𝐸 over 𝑣. Now to figure out 𝐵, we’ll need to know the electric field 𝐸 and the speed of the charge 𝑣.

Considering that electric field 𝐸, if we recall the equation for the electric field of a parallel plate, that electric field 𝐸 is equal to the potential difference divided by the distance separating the plates. This means we can replace 𝐸 in this equation with capital 𝑉 over 𝑑, where 𝑉 is the potential difference given in our problem statement, 𝑑 is the separation between the plates, and 𝑣 is the as yet unknown speed of our particle.

To solve for that speed 𝑣, let’s recall that we’re told the kinetic energy of our electron. And we can recall that classically the kinetic energy of an object is equal to half its mass times its speed squared. If we take this relationship and rearrange it to solve for 𝑣, we find that 𝑣 is the square root of two times the object’s kinetic energy divided by its mass 𝑚.

So we can insert this expression for 𝑣 into our equation for magnetic field 𝐵. We find that the magnetic field is equal to the potential difference 𝑉 over the distance separating the plates 𝑑 times the square root of two 𝐾𝐸 over 𝑚, where 𝑚 is the mass of the electron. And we’ll treat this mass as exactly 9.1 times 10 to the negative 31st kilograms.

We’re almost ready to plug in and solve for 𝐵. Before we do, let’s convert our kinetic energy, which currently is expressed in units of electron volts, into a value expressed in units of joules. To do that, we’ll multiply its given value in electron volts by a conversion factor, which will give us a final value for the kinetic energy when units of joules.

When we now plug in 𝑉, 𝑑, 𝑚, and 𝐾𝐸 into our equation for 𝐵, we’re careful to write our distance values in units of metres to keep units across this entire expression consistent.

When we enter these values on our calculator and solve for 𝐵, we find that to three significant figures it’s 1.13 times 10 to negative third tesla. That’s the magnetic field’s strength acting on the electron.

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