A singly charged ion of mass m is accelerated from rest by a...

A singly charged ion of mass $m$ is accelerated from rest by a potential difference $\Delta V$ . It is then deflected by a uniform magnetic field (perpendicular to the ion's velocity) into a semicircle of radius $R .$ Now a doubly charged ion of mass $m^{\prime}$ is accelerated through the same potential difference and deflected by the same magnetic field into a semicircle of radius $R^{\prime}=2 R .$ What is the ratio of the masses of the ions?

Key Concepts

Acceleration by a Potential Difference

When a charged particle is accelerated from rest by an electric potential difference, the work done by the electric field is converted into kinetic energy. This principle is expressed by equating the work done (charge multiplied by potential difference) to the kinetic energy (one half times mass times velocity squared), allowing us to determine the particle's speed after acceleration.

Magnetic Force and Circular Motion

A charged particle moving perpendicular to a magnetic field experiences a Lorentz force that is always perpendicular to its velocity. This force acts as the centripetal force, causing the particle to move in a circular path. The relationship between the centripetal force (mass times velocity squared divided by radius) and the magnetic force (charge times velocity times magnetic field strength) defines the radius of the circular motion.

Relationship Between Kinetic Energy, Velocity, and Particle Properties

By combining the energy gained from acceleration through a potential difference with the circular motion dynamics in a magnetic field, one can relate the particle's mass and charge to the radius of its trajectory. This relationship is fundamental for analyzing problems where variations in mass or charge affect the path of the particle in a magnetic field.

Transcript

00:01 All righty, so we have a singly charged particle and it's in motion and it's accelerated through a certain region, a certain region of space where there's a potential difference delta v.

00:12 It's got a mass m.

00:13 Here it is.

00:15 And this little vector here is just to try to show you guys it's in motion and this direction here to the right.

00:20 And it's deflected by a uniform magnetic field.

00:23 And when it hits this magnetic field, the particle is then going to make a semicircular kind of a circular.

00:29 Kind of a circle, a semicircle.

00:33 So after it hits the magnetic field, the particle makes a semicircle.

00:37 And it rides this semicircle because of the influence of the magnetic field.

00:41 All right? and we know that the radius of this is r.

00:46 Okay.

00:48 But in a different situation, we have the same potential difference.

00:54 We have a different map stone now.

00:57 I want to call m prime.

00:59 Okay.

01:00 In motion.

01:01 And goes to a magnetic.

01:02 Field and then it goes through a semi -circular arc just like we had before but this time it's twice the size all right excuse my poor drawing of an arc here but you guys get the the idea here and this is the radius here my radial vector here are prime and it's equal to to r all right so the question is we want to figure out what is the ratio of the masses of the ions okay you have these two particles here, these two ions, we want to know what is the ratio or the masses.

01:37 What do i mean by that? at the end, we just want to basically have a m prime over m, that ratio.

01:43 We want to get some sort of number at the end, that's our goal.

01:46 So we're going to go through all the math here and do our best to get to a solution.

01:51 So give ourselves some more space here.

01:53 First thing i'm going to do is this is an isolated system for energy.

01:56 So we're going to go ahead and apply conservation of energy.

02:02 The change of kinetic energy plus any change of potential energy is going to be equal to zero we're going to go ahead and plug in what we know here we can just say k final minus k initial and group those guys together plus the delta u which is equal to zero and we can go ahead and substitute the kinetic energies the actual legitimate term for one -a -half mv final squared minus the one -half m v initial squared those guys together and we're going to replace the delta u with q delta v by definition equal to zero that looks like it's equal to one oh give me one second let's have some more space here equal to zero right below it there cool obviously rearranges pretty nicely we know that our velocity initially is zero so this term right up here goes to zero not to worry about that guy we go down again get myself some more space so we're going to lose that term...

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