A solenoid 80.0 centimeters long is wound with 600 turns of wire. The cross-sectional area of the coil is 6.00 centimeters squared. What is the self-inductance of the solenoid?
We can call the length of the solenoid, 80.0 centimeters, 𝑙. And the number of turns of wire, 600, we’ll call capital 𝑁. The cross-sectional area of the solenoid, 6.00 centimeters squared, we’ll call capital 𝐴. We want to solve for the self-inductance of the solenoid, which we’ll call 𝐿.
To figure out 𝐿, we can recall the mathematical relationship for a solenoid’s self-inductance. This relationship tells us that a solenoid’s self-inductance is equal to the permeability of free space 𝜇 naught times the number of turns in the coil squared times its cross-sectional area 𝐴 over its length.
The constant 𝜇 naught we’ll treat as having an exact value of 1.26 times 10 to the negative sixth tesla meters per amp. When we solve for 𝐿 using our values, each one in the equation is given in our problem statement, or it’s a known constant.
So we’re now ready to plug in. When we do, we’re careful to convert our area into units of meters squared and our length 𝑙 into units of meters. When we enter these values on our calculator, we find that, to three significant figures, 𝐿 is 3.40 times 10 to the negative fourth henries. This is the self-inductance of this solenoid.