(III) A proton (mass mp), a deuteron (m = 2mp , Q = e), and...

(III) A proton (mass $m_p$), a deuteron ($m = 2m_p , Q = e$), and an alpha particle (m = 4m_p , Q = 2_e$) are accelerated by the same potential difference V and then enter a uniform magnetic field $\overrightarrow{B}$ where they move in circular paths perpendicular to $\overrightarrow{B}$ Determine the radius of the paths for the deuteron and alpha particle in terms of that for the proton.

Key Concepts

Energy Conversion via Potential Difference

When a charged particle is accelerated by a potential difference, its electric potential energy (given by qV) is entirely transformed into kinetic energy (½mv²). This conversion allows one to determine the particle's speed immediately after acceleration, using the relation v = ?(2qV/m), which is crucial for analyzing its subsequent motion in a magnetic field.

Lorentz Force and Circular Motion

A charged particle moving in a uniform magnetic field experiences the Lorentz force, which is always perpendicular to its velocity, resulting in circular motion. This force, expressed mathematically as qvB, acts as the necessary centripetal force that keeps the particle moving in a curved path, making it possible to relate the particle's speed, charge, mass, and the magnetic field strength to the radius of the circular path.

Radius of Curvature in a Magnetic Field

The radius of the circular path of a charged particle in a magnetic field is given by r = (mv)/(qB). Incorporating the speed obtained from the energy conversion (v = ?(2qV/m)) into this equation shows that the radius depends on the square root of the mass and is inversely proportional to the charge. Understanding this relationship is key to comparing the trajectories of different particles under identical accelerating conditions.

Transcript

00:01 So this problem can be a little tricky, but the way that we're going to approach question number 20 is to use energy conservation laws.

00:12 So we have a proton, a duteron, and an alpha particle that are all accelerated by some potential difference v, and they enter into a uniform magnetic field where they start to, rotate.

00:32 They each have a circular path perpendicular to be.

00:36 So you can think of this as we have our charged particle accelerating until it hits the magical box of magnetic field, in which case they will start to turn and then they'll have some radius.

00:57 Now if you have a hard cutoff here, then it is just going to come straight out like we saw in another problem.

01:07 But if we just think of it as a circle with some radius r, it's a little bit easier for us to consider.

01:17 So we want to find the radius of the paths for the duteron and alpha particle in terms of that for the proton, which is why i've written the masses.

01:26 So this is the, oh no, it's too high up.

01:31 This is the mass table and then this is the big q charge table.

01:39 And like i hinted at earlier, the way we can solve this is by using energy conservation laws.

01:46 So that our initial energy must always be equal to our final energy no matter what.

01:54 And the reason this helps us here is because the energy of a particle in a potential is going to be converted entirely into kinetic energy.

02:16 So we can, let me be very clear.

02:20 This is a big v.

02:21 This is a potential.

02:24 And then this little v is a velocity.

02:30 And this gives us a relationship to the velocity, which is the square root of 2 times q times big v.

02:44 All divided by the mass of, can be the proton in this case.

02:51 And then there is going to be a force that is turned on right here whenever it enters into the magnetic field.

03:01 And that force, which i'll do in blue, is the simple qvb...

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