Net force between two conducting strips

  • #1
Bling Fizikst
69
7
Homework Statement
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Relevant Equations
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1717650939250.png


I have a lot of doubts regarding this . Let's say the resistance is ##R## where the net interaction force becomes zero . So , the current flowing should be : $$i=\frac{V_{\circ}}{R}$$ Will charges be induced like in capacitors? If yes then let's say ##\pm q## is induced and the length of the strip is ##l##. The corresponding electrostatic force should be : $$F_e=\frac{q^2}{lb\epsilon_{\circ}}$$ For the magnetic force , i am not sure how to find that . I do know for a thin straight conducting wire , it is : $$F=\frac{\mu_{\circ}i_1 i_2 l }{2\pi d}$$ All in all , i am unable to see a clear path .
 
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  • #2
I agree with your electrostatic force. You need to calculate q, though.
For the magnetic force, you could divide each strip into narrower strips and find the horizontal components of the pairwise attractions.
 
  • #3
1717666830887.png


I am pretty sure that ##z=x+r\sin\varphi## when differentiated will give ##dz=dx+r\cos\varphi\cdot d\varphi## Doing all the calculations , i got $$B=\frac{\mu_{\circ} i}{4\pi d}\left(\frac{4}{3}+\frac{\pi}{2}\right)$$
 
  • #4
Assume the conducting plates have negligible resistance.

Convince yourself that the circuit is simply a capacitor and resistor in parallel, with a fixed voltage applied. You should now be able to find the charge and hence the electric force.

As for the magnetic force, I suspect you are not required to derive an expression for it - which would be very messy!

But you can use proportionality. If the current changes by some factor, and the geometry is fixed, by what factor does the magnetic force change?
 
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  • #5
For the RC circuit , let's say total charge is constant ##q=q_1+q_2## such that ##q_1## enters the capacitor and the current ##i_2=\dot{q_2}## enters the resistor . Using kirschoff's law , we can deduce that : $$q_1=CV_{\circ}$$ $$i_2=\frac{V_{\circ}}{R}$$ $$i_2 R-\frac{q_1}{C}=0$$ In the 3rd equation : we can use ##-\dot{q_1}=\dot{q_2}## to get : $$q_1=CV_{\circ}e^{-\frac{t}{\tau}}$$ where ##\tau=RC## . For ##q_2## , $$\dot{q_2}=i_2=\frac{V_{\circ}}{R}\implies q_2=\frac{V_{\circ}t}{R}$$ I think i am messing up somehwere here as well .

For the Magnetic force , ##F_{b}=F_{e}=\eta^2F_{b}'\implies F_{b}'=\frac{F_e}{\eta^2}?## where ##F_{b}'## is the final magnetic force after resistance is changed from ##R\rightarrow \eta R##
 
  • #6
Wasted my time on this . Force on each plate ##F_e=\frac{q^2}{2A\epsilon_{\circ}}## where ##q=CV_{\circ}## and ##C=\frac{lb\epsilon_{\circ}}{d}## $$F_e=F_b= \frac{lb\epsilon_{\circ} V_{\circ}^2}{2d^2}$$ Since ##F_e## remains constant : $$F_b'=\frac{F_e}{\eta^2}$$ So , Net force of interaction per unit length should be : $$\frac{1}{l}\cdot \left(F_e-F_b'\right)=\frac{F_e}{l}\left(1-\frac{1}{\eta^2}\right)=\frac{b\epsilon_{\circ} V_{\circ}^2}{2d^2}\left(1-\frac{1}{\eta^2}\right)$$
 
  • #7
Bling Fizikst said:
For the RC circuit , let's say total charge is constant ##q=q_1+q_2## such that ##q_1## enters the capacitor and the current ##i_2=\dot{q_2}## enters the resistor . Using kirschoff's law , we can deduce that : $$q_1=CV_{\circ}$$ $$i_2=\frac{V_{\circ}}{R}$$ $$i_2 R-\frac{q_1}{C}=0$$ In the 3rd equation : we can use ##-\dot{q_1}=\dot{q_2}## to get : $$q_1=CV_{\circ}e^{-\frac{t}{\tau}}$$ where ##\tau=RC## . For ##q_2## , $$\dot{q_2}=i_2=\frac{V_{\circ}}{R}\implies q_2=\frac{V_{\circ}t}{R}$$ I think i am messing up somehwere here as well .

For the Magnetic force , ##F_{b}=F_{e}=\eta^2F_{b}'\implies F_{b}'=\frac{F_e}{\eta^2}?## where ##F_{b}'## is the final magnetic force after resistance is changed from ##R\rightarrow \eta R##
It looks like you are meant to assume the capacitor is already fully-charged.

Therefore you have:
- a constant current, ##i=\frac {V_0}R##, through the resistor and
- a constant charge ##q=CV_0## in the capacitor.

Forget about the magnetic force for the moment. I suggest first finding expressions for the capacitance, the charge and the electric force. Then find the actual value of the electric force.

Then consider the Post #4 hint about using proportionality.
 

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