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Could someone please explain to me why we want to take the "magnitude" of the emf?

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2 Answers 2

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Just reinforcing @jak and adding my own two pence here really.

I think it is because you are not trying to determine the motion of the current, therefore you don't need it. I also think you don't have enough information to determine the sign.

There is no N number of coils in this, but I think the texas note has a nice example of an induced emf problem, which may help you visualize what is going on. Though is is a bit different to your question. Essentially the direction of the emf tells you whether the current is clockwise or anti-clockwise round a coil.

The texas notes have helped me with a few emag problems so it's worth having a look at them generally, in my view.

Also, although it is not explicitly asking for magnitude in the text. It does seem to be using the magnitude symbol in the mathematical notation, so that would be them asking for the magnitude of induced emf, right there.

Hope that helps.

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Is it wrong to add a minus sign? –  jip Aug 16 '12 at 0:47
    
Generally or for your question? Your answer should be in the form $|\varepsilon|=6.28mV$. –  Magpie Aug 16 '12 at 13:06
    
For my question. –  jip Aug 17 '12 at 20:54
    
yes @jak see post above. A modulus sign means it is a magnitude. You cannot have negative magnitude. –  Magpie Aug 18 '12 at 1:07
    
No I meant not giving the answer as the magnitude –  jip Aug 26 '12 at 23:25

To solve induction problems, you first need to define an orientation of the coil, this orientation induces a positive a negative side for any surfaces whose boundary is the coil, then you can compute the magnetic flux and at last the emf. A positive result means the emf runs a current in the positive orientation of the coil. Since you don't know much about the geometry of B, you can't decide if the emf is positive or negative, that is why you are not asked for the sign.

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Could you provide a picture (like a pictorial counterexample)? Having a tough time getting those words into my head –  jip Aug 4 '12 at 20:40
    
@jak, imagine that there is a uniform current density directed towards you from the screen you are looking at as you read this. By the "right hand rule", the emf is positive for a counter clockwise closed path in the plane of your screen. Were the current instead directed in to the screen, the emf would be negative for the same path. But, the magnitudes would be the same. –  Alfred Centauri Aug 4 '12 at 20:47
    
So why do we take the positive emf in the end? –  jip Aug 4 '12 at 21:11
    
@jak, first of all, I apologize, my example is for mmf, not emf. Nonetheless, the principle is the same. In both cases, there is an oriented surface bound by a path along which the integration takes place. Now, to your question, if you're asked for the magnitude, the direction doesn't matter. It's like asking for speed instead of velocity. What precisely is worrying you about this problem? –  Alfred Centauri Aug 4 '12 at 22:54
    
It doesn't ask for "find the magnitude of the induced emf". It just asks for "induced emf" –  jip Aug 5 '12 at 2:17

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