# Question Video: Finding the Current Density of a Proton Beam

A high-energy proton accelerator produces a proton beam with a radius 𝑟 = 0.73 mm. The constant beam current 𝐼 = 5.0 𝜇A. The beam’s charge density 𝑛 = 3.7 × 10¹¹ protons/m³. What is the magnitude of the current density of the beam? What is the magnitude of the drift velocity of the beam? How much time is required for 2.3 × 10¹⁰ protons to be emitted by the accelerator?

06:30

### Video Transcript

A high-energy proton accelerator produces a proton beam with a radius 𝑟 equals 0.73 millimeters. The constant beam current, 𝐼, equals 5.0 microamps. The beam’s charge density, 𝑛, equals 3.7 times 10 to the 11th protons per meter cubed. What is the magnitude of the current density of the beam? What is the magnitude of the drift velocity of the beam? How much time is required for 2.3 times 10 to the 10th protons to be emitted by the accelerator?

In this statement, we’re told the beam radius, 𝑟, the beam current, 𝐼, and the beam’s charge density, 𝑛. We want to know the magnitude of the current density of the beam. We’ll call that 𝐽. Then we want to solve for the drift velocity of the beam. We’ll call that 𝑉 sub 𝑑. And finally we want to solve for how much time is needed for a certain number of protons to be emitted. We’ll call that time 𝑡.

To start on our solution, we can make a sketch of this proton beam. Looking at a section of our proton beam, the beam we’re told has a radius 𝑟, 0.73 millimeters, carries a known current 𝐼, and the beam has a density of protons, 𝑛, of 3.7 times 10 to the 11th protons per meter cubed. First, we wanna solve for the current density, 𝐽, of this beam. And to do that, we recall that 𝐽 is equal to current divided by the area through which or over which that current flows.

In our case, the area over which our current 𝐼 flows is a circular section. To calculate the area of that cross section, we recall that the area of a circle is equal to its radius squared times 𝜋. So 𝐴 in the denominator of our equation is replaced by 𝜋𝑟 squared. We’ve been given the radius of the beam in the problem statement as well as its current, 𝐼. So we’re now ready to plug in and solve for 𝐽. When we do, we take care to write the current in units of amperes and the distance, 𝑟, in units of meters.

When we calculate this fraction, we find that, to two significant figures, 𝐽 is 3.0 amps per meter squared. That’s the current density of this proton beam. And now let’s move on to solving for the drift velocity of the protons in the beam. This is the speed any individual proton might have because of the electric field creating the beam in the first place. To solve for the drift velocity, let’s start by recalling that current, 𝐼, is defined as the amount of charge 𝑄 that passes a particular point and some time interval, 𝑡.

The current we’ve been given in the problem statement 5.0, microamps, means that 5.0 times 10 to the negative sixth coulombs of charge pass a given point in this beam every second. Recall that this is a beam of protons and the charge on a proton can be treated as exactly 1.6 times 10 to the negative 19th coulombs. So we can rewrite this charge in coulombs in terms of number of protons based on the charge of an individual proton, 𝑄 sub 𝑝.

If we divide this fraction by the charge of an individual proton, then the result is a value for current based on a number of protons that pass a given point per second. If that’s how many protons pass a given point every second, we want to find what volume of proton beam this number of protons occupies. If we call that volume 𝑉, then 𝑉 is equal to the number of protons over the number density of protons, 𝑛. Plugging in the given value for 𝑛, we can see that the units of protons will cancel and we’ll be left with units of meters cubed for our volume.

And entering these values on our calculator, we find that the volume taken up by this number of protons is roughly equal to 84.46 cubic meters. Recall we’ve been using one second as our time interval for this process. So if we can figure out the length, 𝐿, of beam occupied by this many protons, then we’ll be able to use the fact that average speed is equal to distance over time to solve for 𝑉𝑑, the drift velocity of the protons. To solve for 𝐿, we’ll take the volume taken up by this number of protons and divide it by the cross-sectional area of the beam or by 𝜋𝑟 squared.

Since we know 𝑉as well as 𝑟, we can plug to solve for the length 𝐿. We’re careful to write our radius in units of meters. And when we calculate this fraction and solve for 𝐿, we find it’s approximately 5.04 times 10 to the seventh meters. That means that for a cross sectional beam area, 𝐴, this number of protons would take up 5.04 times 10 to the seventh meters of length in the beam. We now have our length as well as our time. So we can use these two values to solve for 𝑉 sub 𝑑.

𝑉 sub 𝑑 is equal to 5.04 times 10 to the seventh meters over one second or, to two significant figures, 5.0 times 10 to the seventh meters per second. That’s the drift velocity of the protons in this beam. Finally, we want to solve for 𝑡. The problem statement tells us that 𝑡 is the time it takes for 2.3 times 10 to the 10th protons to be emitted into the beam. We can use our revised expression for the current 𝐼 to help us solve for 𝑡.

The current 𝐼 gives us a number of protons per some amount of time. So if we divide the given number of protons by 𝐼, that division will give us the time needed for that number of protons to pass into the beam. When we plug in for 𝐼 and perform this division, we find that 𝑡, to two significant figures, is 0.74 milliseconds. That’s how long it takes for this number of protons to enter the proton beam.