A solenoid has a ferromagnetic core with 𝑛 equals 1000 turns of conducting coil per metre wound around the core. The current in the solenoid’s coils is 5.0 amperes. The magnetic field inside the solenoid is 2.0 tesla. What is the magnetic susceptibility of the core material?
We’re told in this statement the number of turns in the coil per metre. And we’re also told the current that runs through each one of the coils. We’ll call that capital 𝐼. Further, we’re told the magnetic field strength inside the solenoid — 2.0 teslas. We’ll name that capital 𝐵. We want to solve for the magnetic susceptibility of the material in the core of the solenoid. We’ll call that 𝜇 sub 𝑟.
To solve for the magnetic susceptibility, we can start by sketching our solenoid with its core. Our solenoid we’re told has 𝑛 or 1000 turns per metre. And the current 𝐼 runs through its coils. Thanks to this current, a magnetic field 𝐵 is set up through the core of the solenoid. The material occupying the solenoid’s core has some magnetic susceptibility. That’s what we want to solve for.
To do that, we can recall the mathematical relationship for the magnetic field within a solenoid. This relationship says that that magnetic field is equal to the magnetic susceptibility of the material in the solenoid’s core multiplied by the permeability of free space 𝜇 nought multiplied by the number of turns per unit length of a coil on a solenoid times the current that runs through those coils. So 𝐵 equals 𝜇 sub 𝑟 times 𝜇 sub nought 𝑛𝐼 or 𝜇 sub 𝑟 is equal to 𝐵 over 𝜇 nought 𝑛 times 𝐼.
In our exercise statement, we’re told 𝐵, 𝐼, and 𝑛. And 𝜇 nought, the constant permeability of free space, we’ll treat as exactly 1.26 times 10 to the negative six tesla metres per amp. We are now ready to plug in and solve for 𝜇 sub 𝑟. When we do and enter these values on our calculator, we find that 𝜇 sub 𝑟 to three significant figures is 317. That’s the magnetic susceptibility of the material in the solenoid’s core.