# simple electromagnetism question

by dajugganaut
Tags: electromagnetism, simple
 P: 33 hi, i have a question, if you could help me, that'd be great: why does a smaller diameter coil result in a stronger electromagnet??
 P: 4 Hmmmm...... in school, we learned that the magnetic field in a solenoid is modeled by the following equation B= πο In/L, where u is equal to 4 π (pi)* 10 to the power of -7 I is equal to current n ο (sorry, don't know how to do subscripts) is equal to number of loops and L is equal to the length of the solenoid In all the problems my textbook, the diameter of the solenoid seems to be irrelevent, (it's always given, but you never need it to calculate the strength) but I suppose that since magnetic fields are nonuniform, meaning that they field strength weakens as the distance increases, a greater diameter (greater distance) results in a smaller field. I could be wrong, so don't quote me.
 P: 1,770 Well if you want a correct answer you need to derive the correct equations too. You will recall that $$B = \mu_{0}nI$$ (where n is the number of turns per unit length) is an approximate expression...you get it using Biot-Savart Law or Ampere's Law if you consider the field to be well behaved. This makes your closed integral very easy and you get this linearized expression for a closed pack coil when you can ignore the pitch of the coil (the distance between successive turns). So if you want to study the effect of the diameter you've got to use a proper model to work it out. You can find this in most electromagnetic texts such as Griffiths.
P: 524

## simple electromagnetism question

 So if you want to study the effect of the diameter you've got to use a proper model to work it out. You can find this in most electromagnetic texts such as Griffiths.
I probably wouldn't recommend Griffiths for an introductory high school class in E&M. My naive assumption/guess without using any fancy math, would be that the coils are closer to the center, so the field is stronger there.
 P: 1,770 Okay I didn't exactly know that we're dealing with a high school problem. In that case, yes, I agree with Gza about the math in E/M. So yes, you have the answer here. Griffiths is perhaps the best intro E/M book by the way and this problem is treated there so you might like to have a look at the diagram and the final equation just to satisfy your curiosity about the nature of the field, if you wish to. To help you avoid doing that right now (sometimes reading an advanced book without much background can be dangerous), here's the blurb of it: when you computed the $$B = \mu_{0}nI$$ you made a major assumption to simplify the closed integral that is the meat of Ampere's Law--that the magnetic field is parallel to a line parallel to the axis of the solenoid in the interior and is zero immediately outside the solenoid. Also you neglected the wavy nature of the field outside the solenoid and at the ends (in advanced E/M you will have to take care of these factors and they become more serious in realworld applications). By the time you exhausted these assumptions, you ended up with a linear-looking integral which you could solve with another assumption which is critical to classical E/M: Symmetry. You can think of the non-idealities of a real life solenoid now: (a) no matter how hard you try you cannot possibly have infinitesimally small distance between successive turns, i.e--an ideal close packing is impossible. (b) due to (a) and otherwise, the field is not exactly parallel to the axis of the solenoid...a better idea may be obtained by drawing flux lines which are curved... (c) just outside the solenoid the field just can't be zero as the turns of the coil will behave as infinite packets of rotating charge infinitesimally spaced so that each sets up a complicated magnetic field vector that may even be time variant owing to the resistive dissipation of the coil's material leading to nonlinear time varying currents (by this time you're probably thinking: man, this is hard! Read further) There are other factors too but the gurus of physics at schools (the teachers and the syllabi framers) have kept the excitement in a box called E/M theory to be taught to you when you're a freshman and have done a consistent level of mathematics to understand the ease of the equations which come up during the consideration of the factors above. Just to add to that...introductory physics is good because even though it is full of incorrect assumptions (and therefore equations like these), it makes you rethink about them so that you are (optimally) well prepared to understand the things you read in high school in a better fashion to improve your attempts at considering factors which did not make hell of a difference in school :-D. Enjoy physics... Cheers Vivek

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