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pi-π
pi-π
07:15
@JohnRennie Hey
John Rennie
John Rennie
Hi :-)
It's been a long time. How are you doing these days?
pi-π
pi-π
@JohnRennie Working nowadays. Completed Bachelors. How about you?
John Rennie
John Rennie
I'm still doing the same as always. Enjoying being retired and passing the time by answering physics questions :-)
Are you enjoying work?
pi-π
pi-π
@JohnRennie Yep. I guess people enjoy working in the initial years. Maybe the boredom sets some years later. "Honeymoon Phase" of the work might not be over yet for me.
John Rennie
John Rennie
I found the "honeymoon period" lasted several years, but usually the job changed or I got a new job on that sort of timescale so I enjoyed most of my working career.
My first job was with a large multinational company (Unilever) and it was fairly easy to switch to a different job in the company. So that kept things interesting for me.
Hopefully you'll find the same :-)
pi-π
pi-π
07:23
I see.
user image
@JohnRennie How do we get this table structure in html?
John Rennie
John Rennie
Use the rowspan attribute.
If you Google you'll find lots of examples, or I can write some sample HTML if you want.
pi-π
pi-π
@JohnRennie A simple sample would do.
John Rennie
John Rennie
Give me a couple of mins and I'll write the code now ...
pi-π
pi-π
@JohnRennie Sure.
John Rennie
John Rennie 
<table>
    <tr>
        <td>Plot</td>
        <td>658665</td>
    </tr>
    <tr>
        <td rowspan="4">Foo</td>
        <td>Bar</td>
    </tr>
    <tr>
        <td>Bar</td>
    </tr>
    <tr>
        <td>Bar</td>
    </tr>
    <tr>
        <td>Bar</td>
    </tr>
</table>
@pi-π Like that
pi-π
pi-π
07:32
@JohnRennie Ah. Okay.
John Rennie
John Rennie
Can you see how it works?
pi-π
pi-π
Yes.
John Rennie
John Rennie
OK :-)
 
2 hours later…
Pizza
Pizza
09:45
Hi @JohnRennie :-)
John Rennie
John Rennie
@Pizza Hi :-)
Pizza
Pizza
Are you free?
John Rennie
John Rennie
Ish, I'm helping another student with some programming at the moment.
If you'd like to post your question I can look at it as soon as I'm free.
Pizza
Pizza
Ok, it's always because I don't have the solution, so I want to accept that I did it right or if there is some simpler way to do the exercises, anyway now I'll write everything down
Three parallel straight conductors in the same plane are arranged at a distance d from each other. A square loop with side a, also in the same plane as the wires, is located at a distance of 10 d. The loop has a resistance R. The three wires carry currents I₁, l₂, and l₃, with directions as shown in the figure. Currents I₁ and l₃ are constant over time, while the current I₂ varies with time in an exponential manner: I₂ = I₀ e^(−t/τ). Calculate:

- The magnetic field at the center of the loop due to the three wires at time *t = 0*;
user image
I'm having internet problems :(
Ok for point 1) B = μ0 * I / (2 * π * r)
μ0 = 4 * π * 10⁻⁷ H/m
Magnetic Field due to I1 at a distance of 10d: B1 = μ0 * I1 / (2 * π * 10 * d)
John Rennie
John Rennie
10:12
@Pizza Yes, and at the centre of the square the fields are all normal to the page so they just add.
Pizza
Pizza
I'm typing slowly, sorry :(
Magnetic field due to I2(t) at distance 9d, where I2 varies with time according to I2(t) = I0 * exp(-t / T): B2(t) = μ0 * I2(t) / (2 * π * 9 * d) = μ0 * I0 * exp(-t / T) / (2 * π * 9 * d)
Magnetic field due to I3 at distance 8d: B3 = μ0 * I3 / (2 * π * 8 * d)
This is step 1
John Rennie
John Rennie
The distances from the wires to the loop are 12d, 11d and 10d not 10d, 9d and 8d ...
Pizza
Pizza
Oh yes, sorry, I corrected it.
With these corrections is it correct?
John Rennie
John Rennie
Yes B = μ₀/2𝜋( I₁/12d + I₂/11d + I₃/10d )
Oh wait, sorry.
I₂ is in opposite direction to I₁ and I₃ so its field is in the opposite direction. We need:
B = μ₀/2𝜋( I₁/12d - I₂/11d + I₃/10d )
The sign of I₂ has to be different from I₁ and I₃
Pizza
Pizza
Ah yes
B(t) = μ₀ / (2π) * ( I₁ / 12d - I₀ * exp(-t / T) / 11d + I₃ / 10d )
Right ?
John Rennie
John Rennie
10:24
Yes
Pizza
Pizza
B(0) ≈ -9.70 * 10⁻⁶ T
So this is point 2
John Rennie
John Rennie
I haven't checked the calculation ... but the method is correct
Pizza
Pizza
Ah ok
John Rennie
John Rennie
I need to drop out for a bit to answer the other guy's question ...
Pizza
Pizza
Ah ok
So I'll try to write it all here
f = μ₀ * Iₐ * I_b / (2 * π * r)
f₁₃ = μ₀ * I₁ * I₃ / (2 * π * 2 * d) and f₂₃ = μ₀ * I₀ * I₃ / (2 * π * d)
Thus, the total force per unit length on wire 3 at time t = 0 is the vector sum of f₁₃ and f₂₃, considering the directions of the forces (attraction or repulsion depending on the direction of the
currents
Then for point 3
I use Faraday's law
𝓔 = - dΦ / dt
where Φ Is the magnetic flux through the loop. The total charge
Q flowing in the loop is:
Q = ∫ 𝓔 / R * dt
Φ = B₂(t) * A , where A = a²
Φ(t) = (μ₀ * I₀ * exp(-t / T) / (2 * π * 11 * d)) * a²
I derive Φ(t) with respect to t to obtain:𝓔 = - dΦ / dt = - a² * (μ₀ * I₀ / (2 * π * 11 * d)) * (-1 / T) * exp(-t / T)
Simplifying: 𝓔 = (μ₀ * I₀ * a²) / (2 * π * 11 * d * T) * exp(-t / T)
So Q = ((μ₀ * I₀ * a²) / (2 * π * 11 * d * T * R)) * ∫ exp(-t / T) * dt
Q = ((μ₀ * I₀ * a²) / (2 * π * 11 * d * T * R)) * [-T * exp(-t / T)] (0 to ∞)
Q = (μ₀ * I₀ * a²) / (2 * π * 11 * d * R)
I don't know how clear it is, if so I can try to explain
brainfreeze
brainfreeze
11:13
Hello @JohnRennie, are you free?
John Rennie
John Rennie
Sorry, I'm busy doing a tutorial on Verilog :-(
brainfreeze
brainfreeze
Alright, no problem! I'll ask later then
 
1 hour later…
John Rennie
John Rennie
12:43
@Pizza Yes :-)
Or add the fields first i.e. B = B₁ - B₂ (minus because they are in opposite directions) then the force on wire 3 is just F₃ = BI₃.
And 3 is correct too. The EMF in the loop is d(BA)/dt = A dB₂/dt because only B₂ changes with time.
brainfreeze
brainfreeze
13:24
Hi @JohnRennie, are you free now?
 
3 hours later…
Pizza
Pizza
16:18
@JohnRennie Oh ok, thank you very much, I'm only reading this now!
John Rennie
John Rennie
OK :-)

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