A 150-V battery is connected across two parallel metal plates of area 28.5 cm^2 and separation 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64x10^-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in Fig. E27.29 <IMAGE>. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

Video transcript

Hi everyone. Let's take a look at this practice problem dealing with magnetic fields. So in this problem, a beam of electrons with charge E is equal to negative 1.60 multiplied by 10 to the negative 19 columns. M ss 9.11 multiplied by 10 to the negative 31 kg is accelerated from rest through a potential difference of 2.0 kilovolts and enters a region between two parallel metal plates perpendicular to the electric field. The plates have an area of 35.0 centimeters squared and are separated by 5.0 millimeters with a 200 volt battery connected across them to make sure that the beam of electrons is deflected between the plates. A magnetic field is applied perpendicular to the electric field between the plates determine the magnitude and direction of the magnetic field. Below the question, we're given a diagram of what was described in the problem. We're also give four possible choices as our answers. For choice A, we have 1.5 multiplied by 10 to the negative three tesla into the page. For choice B we have 1.5 multiplied by 10 to the negative three tesla out of the page. For choice C we have 2.0 multiplied by 10 to the negative three Tesla into the page. And for choice D we have 2.0 multiplied by 10 to the negative three Tesla out of the page. Now, the first thing we're gonna do is look at our parallel plate in this diagram and the bottom of the plate is actually connected to the positive terminal of the battery. And so that means the po the um bottom plate is going to be positively charged. That means the top plate is gonna be negatively charged. So in our diagram, that means that the electric field is going to be pointing upwards. So going from the positive plate to the negative plate, and so when our um particle enters in uh into that region, it's going to fill a force due to this electric field. So let's draw a free B diagram of our electron as it goes through the plates. So it's gonna have two forces acted on it. It's gonna have the force due to the electric field. And that's actually going to be pointing downwards. And that's because the charges here are negative. And so the force is going to be in the opposite direction as the electric field. So since the electric field is pointing upwards, the force is going to point downwards. Now, in order for the electrons to be un deflected between the plates, that means that it has to have zero net force in the vertical direction. So that means that are forced due to the our magnetic field, which will label as FB has to be pointing upward. Now, in order for the two to cancel out their magnitudes must be the same. So the magnitude of FE has to be equal to the magnitude of FB. So here we're going to recall our um formula for the fourth due to the uh electro field on a charged particle. And so if you recall that formula, that is gonna be the absolute value of Q multiplied by the magnitude of E. And here I've used the absolute value because we're just looking at the magnitude and this is gonna have to be equal to the force due to the magnetic field. So you recall your formula for that for its magnitude will have the absolute value of Q multiplied by the speed multiplied by the magnetic field magnitude B. And here um we're gonna have to multiply this by the sign of 90 degrees. And the reason why we're going to use 90 degrees because that's gonna be the angle between the velocity vector and the magnetic field vector because we're told in the problem that the magnetic field is perpendicular to the electric field. And in order to get a force, that means that it also has to be perpendicular to the speed. Otherwise it'd be in the same direction as the speed. And so therefore, you'd get zero magnetic force. So lots of things will uh two things will cancel, the cues will cancel out and the sign of 90 degrees is just one left with just E is equal to VB. Now, the question was to define the magnitude of the magnetic field. Someone will solve this for B and so I'll get B is equal to E divided by V. Now, I was not given the magnitude of the electric field nor was I given the speed of the particle as it enters in. So let's look at the speed. First, what I was given in the problem is the charge, the mass and the potential difference that it, that the electron is accelerated um through starting from rest and going to its final speed. So we're actually gonna use that to calculate our speed for this, we're gonna have to use basically conservation of energy. So here our final connect energy, which if you recall your formula for connect energy is gonna be one half MV squared and that's gonna have to be equal to the energy that's being supplied by this potential difference, which in this case is just gonna be the charge, which is just e multiplied by its potential difference, which will label as capital V not. And here we're going to put absolute value signs around um the E and the V knot. And that's because we know that the energy is being supplied has to be positive since our energy is being positive. So here E is actually gonna be negative but we don't know the sign of the potential difference. We don't know if it's positive or negative and that would depend on how it's set up. But we do know that it is supplying a positive energy since we do have um the connect energy going from zero to some higher value. So we just need to solve this equation for V. So we'll have V is equal to, and to solve this, I'll just multiply both sides of the equation by two and divide by M. So and then take the square root, then we'll have the square root of two multiplied by the absolute value of E multiplied by capital V not. And this is divided by M and I have all three of those quantities I have the charge the mass and the potential difference that's being accelerated through. So I have all those quantities. So now I can look at the electric field. Now, what was given in the problem is the voltage of the battery and the separation distance of the plates. So for this, we can recall our relationship between electric potential and electric fields. So here in this case, the magnitude of our electric field E is going to be equal to our potential of the battery, which we'll just call capital V. And that gets divided by our separation distance which is will label as D. So now I can take both my V and their E formulas that we have found and plug them in for our magnetic field formula. So coming back, we'll have B is equal to, for E, we'll have the capital V divided by D and this gets divided by what I have for the speed lowercase V. That's gonna be the square root of two multiplied by the absolute value of E multiplied by capital V. Not divided by M to this point I can plug in values. So I'll have B is equal to, for the potential of the battery, we have the 200 volts and for the separation distance, we have the five millimeters. So I'll take the 200 volts divided by and here you convert the five millimeters into um meters. So I have five and to convert millimeters into meters, I'll just multiply this by 10 to the negative three that gives me meters. The net quantity is divided by the square root of two multiplied by our charge, which was the 1.6 multiplied by 10 to the negative 19 columns that were given the problem of the 1.6 multiplied by 10 to the negative 19 columns. And here have got rid of the negative sign due to the absolute value. Then the potential difference that it was accelerated to that is the to kill a volts. But I'll need to convert that into volts by multiplying by 10 to the three. So I have two multiplied by 10 to the three volts. And then that gets m the, sorry, that gets divided by the mass of the electron which we are also given in the problem that is the 9.11 multiplied by 10 to the negative 31 kg. So everything on the right hand side of that equation is just a number. So I can plug all those values into my calculator and I'll get B is equal to 1.5 multiplied by 10 to the negative three teslas. So here I've just kept two significant figures. Now, the other thing that we need to figure out is the direction of our magnetic field. And for this, we're gonna have to use the right hand rule. And since we're dealing with a negative charge, one way that we can um figure out the direction of the magnetic field is actually instead of using our right hand to use our left hand. So that's a little trick. So normally, um even for the right hand roll, you point your thumb in the direction of the force, your fingers in the direction of the velocity and your fingers are gonna curl in the direction of the magnetic field. So since this is a negative charge, we're gonna use our left hand. So point our thumb in the direction of the force which is going upwards, we point our fingers in the direction of the velocity vector which is going to the right. And when I curl my fingers, they're pointing towards me. This means that the magnetic field is coming out of the page. So we can indicate that with a circle with a.in the middle of that on our diagram. So now we have both the magnitude and the direction. And if I look at my answer choices, this actually corresponds to answer choice B. So just a quick little recap of what we've done here, we start off by equating the electrical force and the magnetic force acting on the charged particle while traveling between the two plates. And this allowed us to write our magnetic field in terms of the electric field and the speed. In order to calculate the electric field, we just had to use our relationship between electric fields and electric potentials and to find the speed, we need to just basically use energy conservation. So I hope that this has been useful and I'll see you in the next video.