Recent content by Tom_Greening

So would that mean the 20J information is not needed for the overall part of calculations since it is converted to heat?

I think I have to go back and read more... I'm reading a book here... energy is lost to the magnetic field...

Is it lost as heat then?

Thankyou for the reply, but I'm still lost myself. What do you mean by the very last sentence? I understand the concept after reading up on inductors that the inductor stores energy where current rises steadily, but due the electromagnetic motive force the inductor is resisting that with an...

The inductor has a work measured in Joules = 1/2Li^2 but I just can't get my head around the time component. The formula says that the amount of work capacity is capable of inductor, but doesn't allow to incorporate the current in time.

Drawing it would be a DC supply connected to an inductor via a switch. I've seen many diagrams. I've read there is a small amount of resistance, but I think I should ignore that. I've seen the formula I = Vb/R * (1-e^-tR/L)

Yes, i figured that. But not sure where to start looking for that. If possible could you post the equation i need to start using. so far I understand the 1J = 1A per second. So 20J = 20A per second. I'm having difficulty associating the Joules in there.

I was so happy with the assistance provide here by NascentOxygen! Great contributor! I have another question: Homework Statement A 20V battery is connected to an inductor with 1000μH that has 1000 turns with a cross section area of 1cm^2 is connected for 0.3 Sec during where 20J energy are...

Ah, got it. Thanks so much for the assistance NascentOxygen

We weren't provided with area or length values. The question is as the lab cover sheet describes. I suspect i could be misunderstanding the formulas... The only other way to do algebraic ratios is define area/length as=> pi*r^2 / pi*2*r, but I don't think that works either. I'm stumped.

I'm not quite sure I get it. I understand that if I let area A = 1 and say length = 1 for 1 loop then for 2 loops the A = 1 and length = 2 isn't it? So ratio would be 1/2? For four loops ratio is 1/4? Isn't that what I did in case b)?

Ah ok! Yes, that is supposed to be N squared. I don't know how to enter the superscript. Thank you for the nudge in the right direction! So let me do this again... In the case of a) The inductance will remain the same at 100μH, due the fact it is still the same area and same loop. But...

Hello, I'm new to this forum. I hope I am submitting this to the right thread. Just started physics and I am looking to get a little assistance Homework Statement A single loop carries 100mA and has an inductance of 100μH Homework Equations For air cored L = μ0 * (N^2*A/length) L = Nø/i...