Force to apply to a loop moving away from a current-carrying wire

[lorenz0]

Homework Statement

A square loop of wire with $N$ turns, side of length $l$ and resistance $R$ is initially at distance $h$ to the right of an infinitely long wire carrying a steady current $i$.
(1) Find the induced current if the loop starts moving away from the wire at constant speed $v$.
(2) Find the force you would have to apply to the loop so that it continues moving at this constant speed $v$.

Relevant Equations

$\mathcal{E}=-\frac{d\Phi(\vec{B})}{dt},\ \vec{F}=i\vec{l}\times\vec{B}$

What I have done:

(1) $\Phi(\vec{B})=\int_{S}\vec{B}\cdot d\vec{S}=-\frac{N\mu_0 il}{2\pi}\int_{s=h}^{s=h+l}\frac{ds}{s}=-\frac{\mu_0iNl}{2\pi}\ln(\frac{h+l}{h})$
so $\mathcal{E}=-\frac{d\phi(\vec{B})}{dt}=-\frac{\mu_0iNl^2v}{2\pi h(h+l)}$ so $i_{ind}=\frac{\mathcal{E}}{R}=-\frac{\mu_0il^2Nv}{2\pi Rh(h+l)}.$

(2) The force one the left side of the square has greater magnitude than that on the right side (since $i$ and $l$ are the same for both sides but $B_{left}=\frac{\mu_0 i}{2\pi s}>B_{right}=\frac{\mu_0 i}{2\pi (s+l)}$) so the net force on the loop is to the left, hence, to make the loop move away at constant speed $v$ I should apply a force to the right which is equal in magnitude to the net force on the loop due to the magnetic field: $F_{applied}=F_{right}-F_{left}=i_{ind}l(B_{right}-B_{left})=-\frac{\mu_0il^2Nv}{2\pi Rh(h+l)}l\frac{\mu_0i}{2\pi}(\frac{1}{h+l+vt}-\frac{1}{h+vt})=\frac{\mu_0^2 i^2l^4Nv}{(2\pi)^2 Rh(h+l)(h+l+vt)(h+vt)}$, where the last two equalities follow from the fact that, if the loop starts moving to the right at constant speed $v$ at time $t=0$, at time $t$ the left side is at distance $h+vt$ from the wire and the right side at distance $h+vt+l$.

Now, my analysis does make sense to me since, intuitively, as the loop is moving away from the wire the magnetic field gets increasingly weaker so the force to be applied to make it move at constant speed $v$ should get weaker too, but the book I got this problem from says that the force applied should be $F=\frac{\mu_0^2i^2l^4Nv}{(2\pi)^2Rh^2(h+l)^2}$.
Why is the force constant? This result agrees with mine at $t=0$ so is it possible that maybe the text is asking for the force to be applied just as the loop starts moving to the right? Thanks

 

[kuruman]

so $\mathcal{E}=-\frac{d\phi(\vec{B})}{dt}=-\frac{\mu_0iNl^2v}{2\pi h(h+l)}$

After you take the derivative, $h$ is a variable that denotes the instantaneous distance to the wire and $v$ is its (constant) rate of change, $v=\frac{dh}{dt}$. So when you write $F=\frac{\mu_0^2i^2l^4Nv}{(2\pi)^2Rh^2(h+l)^2}$, you cannot treat $h$ as a constant. It would have been clearer to use another variable $y$ as the distance to the wire and write the force as
$F=\frac{\mu_0^2i^2l^4Nv}{(2\pi)^2Ry^2(y+l)^2}$ then replace $y=h+vt$ in the expression.

By the way, I did not check your force expression for correctness.

 

[lorenz0]

After you take the derivative, $h$ is a variable that denotes the instantaneous distance to the wire and $v$ is its (constant) rate of change, $v=\frac{dh}{dt}$. So when you write $F=\frac{\mu_0^2i^2l^4Nv}{(2\pi)^2Rh^2(h+l)^2}$, you cannot treat $h$ as a constant. It would have been clearer to use another variable $y$ as the distance to the wire and write the force as
$F=\frac{\mu_0^2i^2l^4Nv}{(2\pi)^2Ry^2(y+l)^2}$ then replace $y=h+vt$ in the expression.

By the way, I did not check your force expression for correctness.

I see, thanks. So, if I understand correctly, if I treated $h$ as a constant (representing the initial loop-wire distance at $t=0$), $F=\frac{\mu_0^2i^2l^4Nv}{(2\pi)^2Rh(h+l)(h+vt)(h+l+vt)}$ would be a correct formulation for the force to be applied at time $t$, right?

 

[kuruman]

I check your old expression and it is correct. Your new expression looks incorrect. If you had followed my suggestion

Write the force as
$F=\frac{\mu_0^2i^2l^4Nv}{(2\pi)^2Ry^2(y+l)^2}$ then replace $y=h+vt$ in the expression.

you would have gotten the correct expression.

 

[lorenz0]

I check your old expression and it is correct. Your new expression looks incorrect. If you had followed my suggestion

you would have gotten the correct expression.

I see, thank you very much.