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1. Homework Statement

__.__**I originally posted this to the advanced homework -section, but maybe this isn't advanced enough to warrant being posted there**__If a mod sees this, feel free to delete the original on the other forum. I'd do it myself, but I can't.__1. Homework Statement

Calculate the mutual inductance for the long, straight conductor and a conducting loop shaped like an isosceles triangle in the picture below, when a = 10cm.

## Homework Equations

Mutual inductance: [itex]M = \frac{N_1\phi _1}{I_2} = \frac{N_2\phi _2}{I_1}[/itex]

Magnetic flux: [itex]\phi = BA[/itex], where

*B*is the magnetic flux density through a loop and

*A*the area of the loop.

Magnetic flux density for a long straight wire(Ampere): [itex]B=\mu H = \frac{\mu I}{2 \pi R}[/itex], where R is the orthogonal distance from the wire.

## The Attempt at a Solution

In order to calculate M, I need to find out the total flux through the triangular loop. We already know the expression for

*B,*however d

*A*is a bit of a mystery.

Looking at the picture, if at a distance

*r*from the vertex of the triangle we mark:

\begin{cases}

\text{Height of top half of triangle} = y(r)\\

\text{Distance from the vertex} = r

\end{cases}

Then by similarity:

[tex]

\frac{1/2 a}{a} = \frac{y(r)}{r} \iff y(r) = \frac{r}{2}

[/tex]

and if we slice the triangle into small slices with height 2y and width dr:

[tex]

dA=2ydr = r dr

[/tex]

Now [tex]d\phi = d(B(r)A(r)) = B(r)dA(r)[/tex]

Therefore

\begin{array}{ll}

d\phi

&= \frac{\mu I r dr}{2 \pi (a+r)}\\

&= \frac{\mu I r dr}{2r \pi (\frac{a}{r}+1)}\\

&= \frac{\mu I dr}{2 \pi (\frac{a}{r}+1)}\\

&= \frac{\mu I dr}{2 \pi (\frac{a}{r}+1)}\\

&= \frac{\mu I}{2 \pi }\frac{dr}{(\frac{a}{r}+1)}

\end{array}

Looking at this it doesn't seem to be possible to integrate this expression with respect to r from 0 to a. My guess is my derivation of dA is wrong. Any help on that front?

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