A solenoid with 25 turns per centimeter carries a current 𝐼. An electron moves within the solenoid along a circular path that has a radius of 2.0 centimeters and is perpendicular to the axis of the solenoid. If the speed of the electron is 2.0 times 10 to the fifth meters per second, what is 𝐼?
Based on the information given to us, we can call the number of turns per centimeter, 25, 𝑛. And we can name the radius of the electron’s circular orbital, 2.0 centimeters, 𝑟. We’ll call the electron’s speed, 2.0 times 10 to the fifth meters per second, 𝑣. We want to solve for the current running through the solenoid, 𝐼.
To get started, let’s draw a sketch of our scenario. In this scenario, we have a solenoid. And we’re told that, within the coil of the solenoid, there is an electron that moves in a circular orbit. If we look down the end of the solenoid, we would see the electron moving in a circular orbit with a radius of 𝑟 given as 2.0 centimeters.
Based on that circular motion, we wanna solve for the magnitude of the current 𝐼 that runs through the solenoid. Let’s first consider what force might be causing the motion of the electron. When current runs through the loops of the solenoid, a magnetic field we can call 𝐵 is set up through the solenoid’s core. This magnetic field exerts a force on charged particles such as our electron.
We can recall that the magnetic force that’s exerted on a particle, 𝐹 sub 𝐵, is equal in magnitude to the charge of the particle 𝑞 multiplied by its speed times the strength of the magnetic field the particle is in.
Recall that we’re also told that the electron not only moves but moves in a circle. That is, it experiences centripetal acceleration caused by a centripetal or center-seeking force, in this case the magnetic force.
In general, the strength of a centripetal force, 𝐹 sub 𝑐, is equal to the mass of the object it acts on times that object’s speed squared over the radius of the circular path the object travels in. Since, in our case, the physical force creating the centripetal motion is the magnetic force, we can set these two equations equal to one another and write that 𝑞𝑣𝐵 is equal to 𝑚𝑣 squared over 𝑟, where 𝑞 and 𝑣 refer to the charge and speed of the electron, 𝑚 refers to its mass, and 𝑟 to the radius of its circular orbit.
We see in this equation that one factor of the speed 𝑣 cancels from both sides. Now here is where we’re going with this equation. We want to solve for the current 𝐼 in the solenoid. That term doesn’t appear anywhere in this equation. But if we can solve for the magnetic field 𝐵, then we can use the equation for the magnetic field within a solenoid to connect the magnetic field strength with 𝐼.
So let’s solve for 𝐵 here. And let that lead us to solving for the current 𝐼. When we rearrange to solve for 𝐵, we find it’s equal to the mass of the electron times its speed divided by its charge times its radius. And because this is a magnetic field created in a solenoid due to a current running through a series of loops, we can set this equation equal to 𝜇 naught 𝑛 times 𝐼.
When we rearrange this equation to solve for 𝐼, the current 𝐼 through the solenoid is equal to the electron’s mass times its speed divided by its charge times the radius of its orbit times the permeability of free space, 𝜇 naught, multiplied by 𝑛, the number of turns per unit length of the solenoid.
Before we can solve for the current 𝐼, let’s write down some of the constants that appear in this equation. The mass of an electron 𝑚 we’ll treat as 9.1 times 10 to the negative 31st kilograms. The charge magnitude of an electron 𝑞 we’ll treat as the magnitude of negative 1.6 times 10 to the negative 19th coulombs. And the permeability of free space 𝜇 naught is a constant we’ll treat as exactly 1.26 times 10 to the negative sixth tesla meters per amp.
Knowing those constants along with the speed of the electron 𝑣, the radius of its orbit 𝑟, and the number of turns per unit length 𝑛, we’re ready to plug in and solve for 𝐼.
When we do, being careful to use units of meters for our radius 𝑟, and enter these values on our calculator, we find that 𝐼, to three significant figures, is equal to 0.018 amperes. That’s the current flowing through this solenoid that would create this electron motion.