# Homework Help: Time-Varying Magnetic Field

1. Oct 19, 2006

### big man

Question:

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

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.

Last edited by a moderator: Apr 22, 2017
2. Oct 19, 2006

### Norman

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.

3. Oct 19, 2006

### big man

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!