We have a number of field lines perpendicular to one loop of wire with an area $A = 10\textrm{ cm}^2$. The magnetic field is$B= 7.2\times10^{-5}\textrm{ T}$.

You turn the loop and the flux decreases with 30%. Calculate the angle at which the turn is turned.

My idea was:

$\sin( \alpha ) = \dfrac{B_n}{B}$. If I continue this way I get $\alpha = 44^\circ$

My book however goes with:

$B_n = B\cos \alpha$, $\cos \alpha = 0.7$. If you continue this way you get $ \alpha = 46^\circ$.

Why is the latter correct? I have no idea at all how this is done.

I will draw a picture to show my confusion:

enter image description here

Why is this incorrect?

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    $\begingroup$ What is the definition of flux? What type of vector product is used? $\endgroup$ – Jerry Schirmer Jan 16 '13 at 22:29
  • $\begingroup$ @JerrySchirmer My textbook on describes it as 'a measure for the number of magnetic field lines through an area', and we haven't had anything about vector products as school as of yet. $\endgroup$ – Ylyk Coitus Jan 16 '13 at 22:36
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    $\begingroup$ In many cases in physics it's easy to tell that either a sine or a cosine is called for. To decide which, try a limiting case. If $\alpha = 0$ (you don't change anything), your formula predicts $B_n = 0$, while the book predicts $B_n = B$. Which makes more sense? $\endgroup$ – user10851 Jan 16 '13 at 23:07
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    $\begingroup$ @ChrisWhite That can come in very handy, thank you! But I would like to know the trigonometric reason for this, which would help me a lot. $\endgroup$ – Ylyk Coitus Jan 16 '13 at 23:10
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    $\begingroup$ Another approach: what fraction of the lines pierce the loop if the loop is in the plane perpendicular to the field? If it is in a plane parallel with the field? That should be a hint as to what the relevant reference angle and trig function should be. $\endgroup$ – Jerry Schirmer Jan 17 '13 at 3:38

It is $\cos \alpha$, if the angle through which the loop is turned wrt to the initial plane of the loop is $\alpha$. So, if the angle between the final plane and initial plane of the loop is $\alpha$, so is the angle between the final normal and initial normal. The final flux is therefore initial flux X $\cos \alpha$, which you can tell in two ways:

  1. The projected area of the loop perpendicular to the magnetic field is $A \cos{\alpha}$
  2. The magnetic field along the direction of the normal in the final position is $B \cos{\alpha}$
  • $\begingroup$ Why the 0.7? What is the logic behind this? I'm guessing it has to do with the 30% decrease but I don't know why. $\endgroup$ – Ylyk Coitus Jan 18 '13 at 19:43
  • $\begingroup$ Also, could you tell me why my drawing is incorrect? $\endgroup$ – Ylyk Coitus Jan 18 '13 at 20:08
  • $\begingroup$ You said that the flux decreases by 30%, which means $(B-B_n)/B=30/100=0.3$. With $B_n=B\cos\alpha$, this would give $1-\cos\alpha=0.3$ and so $\cos\alpha=0.7$...would it not? $\endgroup$ – Vijay Murthy Jan 18 '13 at 20:11
  • $\begingroup$ @VijayMurthy I'm sorry, that makes sense. I however still don't understand why my drawing is incorrect? $\endgroup$ – Ylyk Coitus Jan 18 '13 at 20:12
  • $\begingroup$ Isn't $\alpha$ the angle between $B$ and $B_n$? In you diagram, it is between $B$ and the other side of the triangle (which would correspond to $B_t$, the transverse component). With $\alpha$ as the top angle in the triangle, wont you get $B_n=B\cos\alpha$...? $\endgroup$ – Vijay Murthy Jan 18 '13 at 20:17

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