Two vertical bars are connected at the bottom via resistance......

[PhDeezNutz]

Homework Statement

Two vertical bars are connected at the bottom via resistance $R = 2 \Omega$ and at the top via the battery with $\varepsilon = 1 \text{ V}$ and internal resistance $r = 2 \Omega$. A sliding bar $\text{AB}$ of mass $m = 0.01 \text{ kg}$ and length $\ell = 0.1 \text{ m}$ can move up or down (with no friction). The system is placed in a uniform magnetic field of $B = 1 \text{ T}$ directed out of the page. Find the terminal velocity of the bar in a gravitational field. (SEE FIGURE BELOW)

Relevant Equations

Kirchhoff Voltage Law (KVL)
Kirchhoff Current Law (KCL)
Lorentz Force Law for a wire:
$\vec{F} = I \vec{\ell} \times \vec{B}$ (for this situation $F = I \ell B$)
Induced voltage from changing magnetic flux:
$\varepsilon_{induced} = - \frac{d\, \phi_B}{d \, t}$

Fundamentally I'm struggling with the premise of the problem. Once I get that I think the math of solving the problem will be straight forward. But like I said I think the problem is fundamentally unsound. I hope to be proven wrong.

• initially there is no current in the lower branch of the circuit because $\text{AB}$ is a short circuit. So effectively the second resistor is invincible to the voltage source.
• We can do KVL for the top loop and find the current in $\text{AB}$
• Then apply the Lorentz Force Law to the movable metal bar $\text{AB}$ and find the force is downwards along with gravity (I guess this part is important)
• My confusion lies in the fact that once the bar starts moving is how do we apply KVL and find the current in $\text{AB}$ after the bar starts moving?

I'll show my work so far but in my opinion it has to be wrong (either that or the question itself is wrong). I've attached a picture of my current direction conventions. The voltage sources drawn in blue are to represent the voltages induced by the change in magnetic flux. They aren't batteries.

$I_1 - I_2 - I_3 = 0$
$\varepsilon_{battery} - \varepsilon_{induced} - I_1 r = 0$
$-RI_2 + \varepsilon_{induced} = 0$

$\left| \varepsilon_{induced} \right| = \frac{d \phi_B}{dt} = B \frac{dA}{dt} = B \ell \frac{dw}{dt} = Bv$ (where $v$ is the velocity of the bar)

I'm going to restate the specific values $\varepsilon_{battery} = 1 \text{ V}$, $r = R = 2 \Omega$, $\ell = 0.10 \text{ m}$, $B = 1 \text{ T}$

Plugging all that into the original system

$I_1 - I_2 - I_3 = 0$
$1 - 0.10v - 2I_1 = 0 \Rightarrow I_1 = \left( -0.5 + 0.05v \right)$
$-2I_2 + 0.10v = 0 \Rightarrow I_2 = 0.05v$

We know $I_3 = I_1 - I_2$ so $I_3 = \left( -0.5 + 0.05v \right) - 0.05v = -0.5$

I have trouble reconciling that $I_3$ would be a constant. If $I_3$ is a constant then the force on the movable bar $\text{AB}$ will be constant through the Lorentz Force Law. I though the force would have to change until it matches the gravitational force and from there we would find the initial velocity.

I'm really confused and would be extremely grateful for any help.

 

[PhDeezNutz]

View attachment 353844

Plugging all that into the original system

$I_1 - I_2 - I_3 = 0$
$1 - 0.10v - 2I_1 = 0 \Rightarrow I_1 = \left( -0.5 + 0.05v \right)$
$-2I_2 + 0.10v = 0 \Rightarrow I_2 = 0.05v$

We know $I_3 = I_1 - I_2$ so $I_3 = \left( -0.5 + 0.05v \right) - 0.05v = -0.5$

I think

$1 - 0.10v - 2I_1 = 0 \Rightarrow I_1 = \left( -0.5 + 0.05v \right)$

Should be

$1 + 0.10v - 2I_1 = 0 \Rightarrow I_1 = \left( -0.5 + 0.05v \right)$

So

$I_3 = -0.5 - 0.10v$

 

[kuruman]

There are two major points that you seem to have overlooked.

1. As the bar falls there will motional emf between points A and B. That emf needs to be taken into account before applying KVL to find the current through the bar.
2. Gravity has something to say in all this and your equations are bereft of $g$. Once you find an expression for the current through the bar, draw a free body diagram and then write Newton's second law. It's a second order ODE, but you do not have to solve it because you are only asked for the terminal velocity.

 

[PhDeezNutz]

I thought I did account for the motional emf with changing flux with the term

$B \ell v = (1)(0.10)(v)$ which I included in both KVLs

Obviously from this you can calculate $I_3$ and apply the Lorentz force law to balance $mg$ at terminal velocity.

@kuruman my power is currently out so I’m typing from my cellphone. I want to apologize in advance for possible radio silence. As always you are tremendously helpful.

Edit: the blue voltage sources in my hand drawn pictures are supposed to represent motional emf sources. Phantom voltage sources if you will.

 

[PhDeezNutz]

@kuruman if you worked it out did you by chance get $v_{terminal} = 14.8 \text{ m/s}$ of course downwards

I need to run to the store real quick I'll be back within 10 minutes to type out the details.

Edit: I'll just do it now
The first equation is KCL the latter two are KVL for loops 1 and 2 respectively (see attached picture)
$I_1 - I_2 - I_3 = 0$
$1 - 0.10v - 2I_1 = 0$ (Top Loop or Loop 1)
$-2I_2 - 0.10v = 0$ (Bottom Loop or Loop 2)

The $0.10v$ comes from the following (note $B$ is constant so there is no need for a $dB$)

$\left| \varepsilon_{induced} \right| = B \frac{dA}{dt} = B \frac{d \left(\ell w \right)}{dt}$.

Where $w = \text { width}$ (which is the downward dimension because i've already called the dimension of the moving bar $\ell$ (a constant))

So

$\left| \varepsilon_{induced} \right| = B \frac{dA}{dt} = B\ell \frac{dw}{dt}$

$\frac{dw}{dt}$ is simply the velocity of the middle bar $v$

$\left| \varepsilon_{induced} \right| = B \ell v = (1) (0.10) v = 0.10v$

Solving for $I_1$ and $I_2$ given the aforementioned system before we explained the $0.10v$ part

$I_1 = - 0.5 + 0.05 v$
$I_2 = - 0.05v$

and if

$I_3 = I_1 - I_2$

then we have

$I_3 = -0.5 + 0.10v$

By the right hand hand rule $I_3$ has to be towards the left in order to oppose downward gravity. Otherwise this problem is nonsensical.

$F_B = F_g$

$I_3 \ell B = m g$

$\left( - 0.5 + 0.10v \right) \left( 0.10 \right) \left( 1 \right) = \left(0.01\right) \left( 9.8 \right)$

solving this for $v$

$v = 14.8 \text{ m/s}$

 

[PhDeezNutz]

I definitely drew the induced currents (and therefore the direction of the phantom voltages wrong)

It should be like this (after using the right hand rule and figuring everything out convention wise). It should be this. Again the blue voltage sources represent induced variable emf (because the middle bar is moving).

 

[kuruman]

@kuruman if you worked it out did you by chance get vterminal=14.8 m/s of course downwards

Yes.

 

[PhDeezNutz]

Yes.

Wooo!!

Thank you!

So it really was just a sign/convention error.