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5:42 AM
@Pizza You integrate from t = 0 to t = ∞ and you'll get a finite result.
We need to consider the red strip I've drawn. The field in the strong is B = μI/2𝜋x so the flux is dφ = BA = μI/2𝜋x × d. dx
And the we get the total flux by integrating from the left edge to the right edge i.e. from x to x+a
 
2 hours later…
8:04 AM
$B = \frac{\mu_0 i_0}{2 \pi x}$
$\text{d}\Phi = B \cdot A = \frac{\mu_0 i_0}{2 \pi x} \times d \times \text{d}x$
$\Phi = \int_{x}^{x+a} \frac{\mu_0 i_0 d}{2 \pi x} \, \text{d}x$
$\Phi = \frac{\mu_0 i_0 d}{2 \pi} \int_{x}^{x+a} \frac{1}{x} \, \text{d}x \\
\Phi = \frac{\mu_0 i_0 d}{2 \pi} \ln \left( \frac{x + a}{x} \right)$
@JohnRennie Hi :-), do you happen to mean that?
Yes :-)
Ah but now step 3 is also wrong
$\mathcal{E} = -\frac{\mu_0 i_0 d}{2 \pi} \left( \frac{1}{x + a} - \frac{1}{x} \right) \cdot v$
$I = \frac{\mathcal{E}}{R} = \frac{\mu_0 i_0 d v}{2 \pi R} \left( \frac{1}{x} - \frac{1}{x + a} \right)$
Are these correct now?
anyway, can you see the LaTeX or is it still giving problems?
I'm busy at the moment. Sorry :-(
8:26 AM
It's okay, don't worry
 
2 hours later…
10:02 AM
@Pizza Yes I get the same result for I.
@JohnRennie ah ok,so now I just need to calculate the total charge?
$Q = \int_{0}^{\infty} I(t) \, dt$
Yes
At the moment you have I(x) as a function of x, but it's easy to convert this to a function of time because x = x₀ + vt
Where x₀ is the distance from the wire at time zero.
$x(t) = x_0 + v t$
Yes
$I(t) = \frac{\mu_0 i_0 d v}{2 \pi R} \left( \frac{1}{x_0 + v t} - \frac{1}{x_0 + v t + a} \right)$
$Q = \frac{\mu_0 i_0 d v}{2 \pi R} \int_{0}^{\infty} \left( \frac{1}{x_0 + v t} - \frac{1}{x_0 + v t + a} \right) \, dt$
$Q = \frac{\mu_0 i_0 d v}{2 \pi R} \left[ \int_{0}^{\infty} \frac{1}{x_0 + v t} \, dt - \int_{0}^{\infty} \frac{1}{x_0 + v t + a} \, dt \right]$
So I have to calculate these , right?
Yes, though I'm a bit concerned that you're going to end up with ln(x₀ + vt) and that diverges as x ⟶ ∞
So I'm not sure that the total charge is going to come out as a finite number.
10:26 AM
It shouldn't come out finite i think too

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