Inductance of a Solenoid. (b) A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

Video transcript

Welcome back, everybody. We are making observations about a metallic spiral binding or a coil bind. We're told a couple different things. We're told that in total, there are 48 turns. We are told that it has a diameter of mm and a length of mm. And we are tasked with finding what the self inducted will be if it is connected to a circuit. Luckily, we have a formula for this. It is our constant new not times the number of terms squared times the area of the cross section divided by the length. We know all these terms except the area. So let's go ahead and calculate the area real quick. We have that the area of the cross section is just gonna be the area of a circle. So this will be pi times radius squared. So this will be pie times well, we're gonna take our diameter and divide it by two. We'll have 25 divided by two, but then we need it in meters. So we're gonna multiply the whole thing by 10 to the negative third and then we're going to square that value multiplying straight across, we get an area of 4.91 times 10 to the negative fourth meters squared. Great. So now we are ready to find the self conductance. So let's go ahead and do that if you not, which is equal to four pi times 10 to the negative seven times 48 squared times R area of 4.91 times 10 to the negative fourth all divided by our length of 297 millimeters. But we want this in meters. So I'm gonna multiply this by 10 to the negative third. This gives us a final answer of 4. micro Hertz which corresponds to our final answer. Choice of a Thank you all so much for watching. Hope this video helped. We'll see you all in the next one.