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I am working on a physics problem where I am told that there are two questions:

EQ1:

$$ W = \left(\frac{\mu_0}{2\pi}\right)\left(\frac{L}{R}\right)I_1^2+I_1LB_{e,z} $$

EQ2:

$$ W = \left(\frac{\mu_0}{2\pi}\right)\left(\frac{L}{R}\right)I_2^2-I_2LB_{e,z} $$

Supposedly, if you subtract E2 from E1 it equals:

$$ B_{e,z} = \left(\frac{\mu_0}{2\pi R}\right)(I_2-I_1) $$

How is this possible? Can someone explain this to me?

mzp
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AAA
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  • @mzp The equations you have written do not match the equations that were in the photos in the original post. The second equation was $W=\left(\frac{\mu_0}{2\pi} \right)\left(\frac{L}{R}\right) I_2^2 - I_2L B_{e,z}$. – kccu Dec 03 '15 at 01:48
  • @kccu Oh, I see. Sorry about that. I'll fix it. – mzp Dec 03 '15 at 01:49
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    If you’re uncomfortable with the idea of subtracting one equation from another, instead set the right-hand sides equal to each other (since they both equal $W$) and solve for $B_{e,z}$. – amd Dec 03 '15 at 07:03

2 Answers2

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Subtracting the two equations you get

\begin{align} W -W &= \left(\frac{\mu_0}{2\pi}\right)\left(\frac{L}{R}\right)I_1^2+I_1LB_{e,z} -\left(\frac{\mu_0}{2\pi}\right)\left(\frac{L}{R}\right)I_2^2+I_2LB_{e,z} \\[2ex] 0 &= \left(\frac{\mu_0}{2\pi}\right)\left(\frac{L}{R}\right)(I_1^2-I_2^2)+(I_1+I_2)LB_{e,z} \\[2ex] -(I_1+I_2)LB_{e,z} &= \left(\frac{\mu_0}{2\pi}\right)\left(\frac{L}{R}\right)(I_1^2-I_2^2) \\[2ex] (I_1+I_2)B_{e,z} &= \left(\frac{\mu_0}{2\pi}\right)\left(\frac{1}{R}\right)(I_2^2-I_1^2) \\[2ex] B_{e,z} &= \left(\frac{\mu_0}{2\pi R}\right)\frac{(I_2^2-I_1^2)}{(I_1+I_2)} \\[2ex] \end{align}

Finally, notice that

\begin{align} \frac{(I_2^2-I_1^2)}{(I_1+I_2)} &= \frac{(I_2-I_1)(I_1+I_2)}{(I_2+I_1)}= (I_1-I_2). \end{align}

mzp
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  • You should have $(I_1+I_2)$ everywhere you have $(I_1-I_2)$, and the first line should have a plus sign before the last term. – kccu Dec 03 '15 at 01:46
  • See my comment on the original question. You edited the photos out of the equation, but copied them incorrectly. (Also notice that your last line is incorrect - you would have $I_1+I_2$, which is not what was to be shown.) – kccu Dec 03 '15 at 01:49
  • THANK YOU both! I appreciate the help - you have cleared up a lot of confusion for me. – AAA Dec 03 '15 at 01:56
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Subtracting the second equation from the first, \begin{align*} W-W &= \left(\frac{\mu_0}{2\pi}\right) \left(\frac{L}{R}\right) I_1^2+I_1LB_{e,z} - \left(\left(\frac{\mu_0}{2\pi}\right) \left(\frac{L}{R}\right) I_2^2-I_2LB_{e,z}\right)\\ 0 &= \left(\frac{\mu_0}{2\pi}\right) \left(\frac{L}{R}\right) I_1^2- \left(\frac{\mu_0}{2\pi}\right) \left(\frac{L}{R}\right) I_2^2+(I_1LB_{e,z}+I_2LB_{e,z} )\\ -(I_1LB_{e,z}+I_2LB_{e,z})&=\left(\frac{\mu_0}{2\pi}\right) \left(\frac{L}{R}\right) I_1^2-\left(\frac{\mu_0}{2\pi}\right) \left(\frac{L}{R}\right) I_2^2\\ -B_{e,z}(I_1+I_2)L &= \frac{\mu_0}{2 \pi R}(I_1^2-I_2^2) L \\ -B_{e,z}(I_1+I_2)L &= \frac{\mu_0}{2 \pi R}(I_1-I_2)(I_1+I_2) L \end{align*} Assuming $L \neq 0$ and $I_1+I_2 \neq 0$ we can cancel those on both sides to get $-B_{e,z} = \frac{\mu_0}{2\pi R}(I_1-I_2)$, so $B_{e,z}=\frac{\mu_0}{2\pi R}(I_2-I_1)$.

kccu
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