A solenoid is wound with 2000 turns per meter. When the current is 5.2 amps, what is the magnetic field within the solenoid?
We can call the number of turns per meter, 2000, 𝑁. And we can call the current that runs through the solenoid, 5.2 amperes, 𝐼. We want to solve for the magnetic field that’s inside the solenoid. We’ll call that field 𝐵.
If we make a sketch of our solenoid, we see it’s a coil of wire with a current 𝐼 running through each loop. The current running through these loops creates a magnetic field that exists inside the solenoid that we’ve called 𝐵.
There is a mathematical relationship for the magnetic field inside a solenoid that we can recall. This relationship says that the magnetic field in the solenoid is equal to the permeability of free space, 𝜇 naught, times the number of turns in the solenoid multiplied by the current that runs through each turn.
𝜇 naught is a constant whose value we’ll assume is exactly 1.26 times 10 to the negative sixth tesla-meters per amp. Knowing 𝜇 naught, 𝑁, and 𝐼, we’re now ready to plug in and solve for 𝐵. Before we do, just note briefly that 𝑁 we’re told is the number of turns per meter.
So the magnetic field we’re solving for is the magnetic field for a meter-long solenoid. When we enter these values on our calculator, we find that, to two significant figures, 𝐵 is equal to 1.3 times 10 to the negative two tesla. That’s the magnitude of the magnetic field within the solenoid.