Question Video: Determining the Orbital Radii of Charged Particles in a Magnetic Field

Two particles have the same linear momentum, but particle 𝐴 has four times the charge particle 𝐡. If both particles move in a plane perpendicular to a uniform magnetic field, what is the ratio (𝑅_(𝐴))/(𝑅_(𝐡)) of the radii of their circular orbits?

03:15

Video Transcript

Two particles have the same linear momentum, but particle 𝐴 has four times the charge particle 𝐡. If both particles move in a plane perpendicular to a uniform magnetic field, what is the ratio 𝑅 sub 𝐴 to 𝑅 sub 𝐡 of the radii of their circular orbits?

To start off, let’s consider the scenario of what’s happening. We have two particles, we can call the blue one particle 𝐴 and the pink one particle 𝐡, entering a magnetic field of uniform strength that we’ve called capital 𝐡. We’re told in the problem statement that these particles have the same linear momentum, which means that if we take the product of their mass and their velocity we get the same number. This means that if we write that 𝑝 linear momentum is equal to π‘š times 𝑣, then we can say that 𝑝 sub 𝐴, the linear momentum of particle 𝐴, is equal to 𝑝 sub 𝐡, that of particle 𝐡.

But not only do the particles have linear momentum, they also are charged particles; they have a nonelectric charge to them. That means that as the particles enter this uniform magnetic field, they’ll experience a magnetic force which will push them in a circular path. There’s a very tidy equation for the radius of that circular path that a charged particle follows in the uniform magnetic field. The radius of that circular arc is known to equal the linear momentum of the particle π‘š times 𝑣 divided by its charge times the strength of the magnetic field that it’s in. What we want to solve for is the ratio of those radii for particle 𝐴 and particle 𝐡. As we do this, we saw that the linear momentum the two particles is equal, and we’re also told in the problem statement that the charge of particle 𝐴 is equal to four times the charge of particle 𝐡.

Keeping these two relationships in mind, let’s now write expressions for the circular radii of particle 𝐴 and particle 𝐡 separately. The circular radius of particle 𝐴 is equal to 𝑝 sub 𝐴, its linear momentum, divided by its charge times capital 𝐡, the magnetic field that this particle is in. And then the radius of particle 𝐡 is equal to 𝑝 sub 𝐡 divided by 𝑄 sub 𝐡 times capital 𝐡, which is not the particle 𝐡 but rather the magnetic field. Now that we have these expressions, let’s divide them one by another to find the ratio 𝑅 sub 𝐴 to 𝑅 sub 𝐡. That equation then for 𝑅 sub 𝐴 to 𝑅 sub 𝐡 is equal to this expression. And we see right away that the magnetic field, capital 𝐡, cancels from these terms; it divides out.

At this point we can start to use the information given to us in the problem statement. For example, 𝑝 sub 𝐴, the linear momentum of particle 𝐴, is equal to that of particle 𝐡. This means that we can replace 𝑝 sub 𝐡 in our denominator with 𝑝 sub 𝐴. And note that we can also just as well have replaced 𝑝 sub 𝐴 with 𝑝 sub 𝐡. Regardless of how we do that substitution, we see that this term now also cancels from our expression. We can then simplify the right-hand side of this expression so that it becomes 𝑄 sub 𝐡 over 𝑄 sub 𝐴. That’s what 𝑅 sub 𝐴 over 𝑅 sub 𝐡 is equal to. But now we can make yet another substitution because 𝑄 sub 𝐴 we know is equal to four times 𝑄 sub 𝐡. And with that substitution made, we see that the factor of 𝑄 sub 𝐡 cancels out. We’re left then with a purely numerical ratio: one to four. That then is our answer. That’s the ratio of the radius of particle 𝐴 to the radius of particle 𝐡 in this magnetic field.

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