Skip to main content
Ch 27: Magnetic Field and Magnetic Forces

Chapter 27, Problem 27

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?

Verified Solution
Video duration:
6m
This video solution was recommended by our tutors as helpful for the problem above.
1029
views
Was this helpful?

Video transcript

Hey everyone. So in this video we are working with magnetic and electric fields. Alright, let's see what they are giving us and what they're asking from us. So we have a velocity selector in a mass spectrometer uses a magnetic field of a given strength and an electric field of a given strength. The fields are uniform and perpendicular. We have a doubly ionized gold ions of a given mass passed through the velocity selector zone with an initial velocity that's perpendicular to both of the fields. They're asking us to determine the speed of the particles that passed through without being deflected. And then they tell us that the particles pass through. Another reason, another region they travel in a circular trajectory with a given radius and they're asking us to determine that mass. So the first thing we need to recall is that crossed electric and magnetic fields can be used as a velocity sector and they cancel when the force is equal each other. So when the force from the magnetic field equals the force from the electric field. So from there we can recall that the force of the magnetic field is given as Q. B. E. Times the sine of theta. And the force from an electric field is given as Q. E. So in this problem they tell us that um the so the particles and the magnetic fields are perpendicular to each other. And so that data is 90 so that sino nineties one. So when we set the um forces equal to each other, we get Q. B. B equals Q. E. The charges cancel. And now we are left to solve for V equals E over B. They give us both of those values in the problem. And so our speed is our electric field 92. That was given to us as .31 times 10 to the 4th volts per meter over our magnetic field magnitude. Which is given to us 0. Tesla. And we can recall that Tesla are just volt seconds per meter squared when we rewrite that um unit in terms of volts. And so that makes our unit conversion very easy there. And so we kind of plug that into our calculators and we get two point oh seven times 10 to the fourth meters per second. And so we look at our potential answers and so far so the only correct answer for part one is C. But we're still going to solve out For part two. Just find the mask, make sure that we're on the right track. Okay, so for the mass we can recall that the radius in a magnetic field is given as our equals N. V over Q B. Q is actually the absolute value of Q. There. So let's take each of those terms one by one. Our was given to us in the problem. So it was 14 centimeters. Whenever you write that as 140.14 m V. We just solved for 2.07 times 10 to the 4th meters per second Q. Is we are told in the problem that we have a doubly ionized adam or particle. And so that's the charge is two plus And so that tells us that the charge is two times 1.6 times 10 to the minus 19 C. That's just our electric um constant, their electric charge constant that we can recall. And so R. B magnetic field term is .15. Tesla is again given to us in the problem and from there it is just a plug and chug. We can rearrange this equation in terms of mass because that's what we're solving for. So that would be our cue be over be and we'll just plug each of those terms and that's what we know all of that over the speed that we just found part one And we are left with a mass of 3.25 times 10 to the - kg. And if we go back up to our answer choices, we will see that. Yes. The answer for part two is correct. In C. So that confirms that C is the correct answer for this problem. That's all we have for this one. We'll see you in the next video
Related Practice