## Time-Varying Magnetic Field

Question:

The link below gives the question I'm interested in. The question is p6.2.

http://www.tkk.fi/Yksikot/Sahkomagne...1_2006_h07.pdf

My Thoughts

OK, now I thought I knew how to do this question. The expression for the magnetic field shows that it varies with the position along y and varies with time.

So what I thought you'd do is this:

$$\Phi = \int{B.ds}$$

Where the differential area is given as:

$$dS_z=dx dy$$

So the above integral becomes:

$$\Phi = b B_0 \int{B.dy}$$

where b is dx (0.1) and $$B_0=3 \mu T$$ and B is the rest of the expression for the varying magnetic field.

So I thought you'd integrate this expression with limits of 0 to 0.3, but the answer I get is nothing like their answer??

I end up with an expression for the current i of:

$$i=- \frac {B_0 b \omega} {R k}[ cos(\omega t -k a)-cos(\omega t)]$$

This is nothing like their expression of:

$$i=- \frac {B_0 b \omega} {R k} [sin(\frac {1} {2} k a) sin(\omega t- \frac {1} {2} k a)]$$

This is simplified by them in the hints section (just scroll down the page to find the hints).

So what am I doing wrong here??

EDIT: oops my expressions for I are meant to contain cosines and not sines.
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 Blog Entries: 12 Try using this obscure trig property: $$sin(x) - sin(y) = 2 sin(\frac{x-y}{2})cos(\frac{x+y}{2})$$ See if that helps, or throw some numbers in and graph to see if they are the same.
 Thank you so much Norman!!! That works out perfectly in getting the expression that they have. Took me a while to find the trigonometric product formulae. This probably was annoying the crap out of me because I was sure I had it right and it's just such a relief to have it fully completed. Thanks again!