An alpha particle has a charge of 2e and a mass of 664 times...

An $\alpha$-particle has a charge of $+2e$ and a mass of $6.64 \times 10^{-27}$ kg. It is accelerated from rest through a potential difference that has a value of $1.96 \times 10^3$ V and then enters a uniform magnetic field whose magnitude is $3.57$ T. The $\alpha$-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the $\alpha$-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path? (a) Number 1.38E7 Units m/s (b) Number 1.57E-11 Units N (c) Number 0.0802 Units m

Transcript

0:00 Hello.

00:01 In this problem, we're asked to figure out several characteristics of an alpha particle that is going to be accelerated through a potential difference and then travel through a magnetic field completely perpendicularly.

00:14 And we know when that happens, the particle is going to follow our circular motion that we've studied previously and that we know how to define the radius of that motion.

00:24 So let's just think about the situation here and then solve for what we need to.

00:28 So initially, we are asked to solve for for the speed of the alpha particle.

00:33 Well, this alpha particle is going to be accelerated through a potential difference from some higher potential to some lower potential of a particular value.

00:44 I think in this case, it is 1 .2 times 10 to the 6 volts.

00:49 What happens here is we have our charge q, and that's going to be a plus 2e because it's an alpha particle.

00:58 It's going to start at some initial velocity here of 0.

01:01 Travel through this potential difference and then when it gets here is going to be traveling at set new final velocity f f f just some new final velocity that it's accelerated up to let me rewrite that better okay so the first question part a is asking us how fast is this particle going to be traveling well we have to go back and think about accelerating through potential differences from previous chapters but we know that when this charge charge goes to this potential drop is going to lose an amount of potential energy given by the value of the charge multiplied by that potential drop.

01:37 But that loss of potential energy is going to be a gain in kinetic energy of our particle.

01:44 And that equals a one -half mv squared, the final squared technically minus one -half m the initial squared.

01:52 But our initial velocity is zero.

01:54 So if you want to find that final velocity, we just have to figure out what our loss of potential energy is.

02:00 And we know that because we know the charge of our particle and the potential drop that it goes through.

02:04 So let's write this expression here and then solve.

02:08 So we have an equation equating our potential energy loss to our kinetic energy gain.

02:16 And then i just want to solve for my final velocity because i want to know how fast my particle is going to be traveling through that magnetic field so i can tell things about the magnetic force on that particle.

02:25 So solving for this, we see that we get the square root of twice the charge of the charge of our particle multiplied by the potential drop, all divided by the mass of our particle.

02:36 So i plug all these things in.

02:39 Let's see, i get the square root of two times twice the fundamental charge, 1 .6 times 10 to the negative 19 couloms.

02:48 And that's going to all be multiplied in the numerator by our 1 .2 times 10 to the 6 volts.

02:56 Then of course we're all going to divide this here.

02:59 This is a c.

03:00 Divide this here by our mass of a alpha particle which is given to us as 6 .64 times 10 to the negative 27 kilograms.

03:11 Put all this in our calculator and we see for part a we have a final answer that the speed of this particle is about 1 .08 times 10 to the 7 meters per second.

03:21 So again pretty quick but not quite enough to have us have to think about relativity...

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