Magnetic Forces: Deriving Equation to Balance Rider Weight

  • Thread starter Thread starter BOAS
  • Start date Start date
  • Tags Tags
    Forces Magnetic
AI Thread Summary
The discussion focuses on deriving the equation for the magnetic force of mutual repulsion between wires, specifically how it balances the weight of a rider. The equation presented is F = (μ₀ I₁ I₂ L) / (2 π r), which should equal mg, the weight of the rider. The contributor expresses uncertainty about the significance of substituting I² with I₁I₂ and whether their steps adequately demonstrate the cancellation of the Earth's magnetic field. Another participant notes that the problem formulation may be unclear, suggesting that the context of the experiment involving a rider is not well understood. The conversation emphasizes the importance of clarity in experimental setups and the derivation process.
BOAS
Messages
546
Reaction score
19
Hello,

i'm working through some prep required for my lab course and having some trouble understanding an equation. We're asked to show it's derivation and I think I've done this but the script itself is a bit ambigous about how to get there.

Homework Statement



The magnetic force of mutual repulsion between the wires is F = \frac{\mu_{0} I^{2} L}{2 \pi r}

Ideally this force should balance the weight of the rider, however there may also be a contribution from the Earth's magnetic field, which will add a force B_{e} IL

By replacing I^{2} with I_{1}I_{2} the effect of the external fields should cancel.

Homework Equations





The Attempt at a Solution



Depending on the direction of the current, the Earth's magnetic field is either contributing to or opposing the force due to the magnetic field of the wire, so

F_{1} = \frac{\mu_{0} I_{1}I_{2} L}{2 \pi r} + B_{e} IL

F_{2} = \frac{\mu_{0} I_{1}I_{2} L}{2 \pi r} - B_{e} IL

F_{1} + F_{2} = \frac{\mu_{0} I_{1}I_{2} L + \mu_{0} I_{1}I_{2}L}{2 \pi r} (Earth's magnetic field cancels)

2F = \frac{2(\mu_{0} I_{1}I_{2} L) }{2 \pi r}

F = \frac{\mu_{0} I_{1}I_{2} L}{2 \pi r} this is the equation I wanted to reach. Are my steps to get here ok? I don't really know how to put into words the importance of substituting I^2 for I1I2...

Since this force should balance the weight of the rider we can say;

mg = \frac{\mu_{0} I_{1}I_{2} L}{2 \pi r} and rearrange to get

I_{1}I_{2} = (\frac{g2 \pi r}{\mu_{0} L}m Which is what I want to plot a graph of)

Apologies if this is a bit of a vague post, but I don't have much practice of lab work and I don't know if my steps count as showing why the magnetic field of the Earth cancels due to the substitution of I^2 that I made. I do include a few more steps in the algebra in my book, showing explicitly the force due to the Earth's magnetic field cancelling etc.

Thanks,

BOAS
 
Physics news on Phys.org
You're not getting much quick help here. Could it be the problem formulation is somewhat unclear to others ? They may know about forces between current carrying wires, but who is riding them ?
You are to do some experiment, measure something and plot the results in a smart way, right ?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top