- #1

- 148

- 1

## Homework Statement

Hello out there...

I've kinda figured this out, but I'm not quite sure how tbh.

I got this problem:

http://www.gratisupload.dk/download/41959/" [Broken]

The length

*a*is constant, but

*b*varies in time like this:

[tex]\[b\left( t \right)=a\left( 1+{{\left( \frac{t}{\tau } \right)}^{2}}-2{{\left( \frac{t}{\tau } \right)}^{3}} \right),\][/tex]

where [itex]\tau[/itex] is a timeconstant. Besides that I know that for

*t*< 0 then

*b*=

*a*, and for

*t*> [itex]\tau[/itex] then

*b*= 2

*a*.

The magnetic fields produced by the current in the conductors (

*I*

_{1}and

*I*

_{2}) gives a magnetic flux through the rectangular loop of:

[tex]{{\Phi }_{B}}=K\cdot b\left( t \right)[/tex]

Determine the constant

*K*.

## Homework Equations

[tex]B=\frac{{{\mu }_{0}}I}{2\pi r}[/tex]

[tex]d{{\Phi }_{B}}=BdA=\frac{{{\mu }_{0}}I}{2\pi }L\,dr,[/tex]

where

*L*is

*b*(

*t*)

## The Attempt at a Solution

What I've done is as following:

[tex]\[{{\Phi }_{B}}=\int_{a}^{3a}{BdA}=\int_{a}^{3a}{\frac{{{\mu }_{0}}{{I}_{1}}}{2\pi }b\left( t \right)\,dr}+\int_{a}^{3a}{\frac{{{\mu }_{0}}{{I}_{2}}}{2\pi }b\left( t \right)\,dr}=-\frac{a{{\mu }_{0}}\left( {{I}_{1}}+{{I}_{2}} \right)\ln \left( 3 \right)\left( 2{{t}^{3}}-3\tau {{t}^{2}}-{{\tau }^{3}} \right)}{2\pi {{\tau }^{3}}}\][/tex]

Putting this equal the magnetic flux I know, and then solving for K, I get:

[tex]K=\frac{{{\mu }_{0}}\left( {{I}_{1}}+{{I}_{2}} \right)\ln \left( 3 \right)}{2\pi },[/tex]

which supposedly is the correct answer according to my book.

But what I don't understand is, that when I tried the limits

*a*to 2

*a*, which seems more obviously to me, I get ln(2) instead of ln(3). So I don't understand why the limits should be

*a*to 3

*a*instead - if I've even done it correctly in the first place.

So I thought one of you might knew :)

Regards.

Last edited by a moderator: