Faraday's Law - loop turning in the Earth's B-field

[Taulant Sholla]

Homework Statement

Homework Equations

Faraday's Law, Ohm's Law, definition of current[/B]

The Attempt at a Solution

We were given this solution:[/B]

The above solution is leaving out a lot of intermediary steps. I don't agree that "the axis of the coil is at 20°, not 70°, from the magnetic field of the Earth." Technically, the coil's surface normal starts at an angle of θ_i=(90+70)=160°, and ends-up at θ_f=(90-70)=20° I want to produce a totally complete, step-by-step solution. Here's how I'm starting...

I'm not sure how to proceed after #7. How do I evaluate the integral (not sure how to deal with dA as the integration variable with cosθ sitting there??

Any help is appreciated!

 

[gleem]

The first thing I see is that eq. 5 is not correct. dΦ ≠ d(B⋅dA). You missed something in eq. 6 WRT dA. WRT eq 7 reconsider your integration limits.

 

[VKint]

The flux integral should be taken over the entire area of the loop at a given instant in time. In particular, $\theta_i$ and $\theta_f$ are not bounds for the inner (flux) integral--in fact, they're bounds for the outer (time) integral, since $\theta$ is a function of time.

This actually simplifies things quite a bit, because as far as the inner integral is concerned, $\theta$ is a constant--meaning that factor of $\cos(\theta)$ can essentially be ignored when evaluating the integral. The expression $$ \int \cos(\theta) dA $$ then simply becomes $A \cos(\theta)$, where $A$ is the total area of the loop.

I actually like the way you've written the expression in 7 quite a bit, although I haven't seen it this way before. Using this notation, we have: $$ \frac{-NB}{R} \int_{\theta_i}^{\theta_f} d (A \cos(\theta)) = \left( \frac{-NB}{R} \right) A (\cos(\theta_f) - \cos(\theta_i)) $$ Note that this is identical to the expression given in the solution, except for your minor disagreement about the values of $\theta_i$ and $\theta_f$. I'll let you think about why that distinction actually doesn't matter in the end.

 

[Taulant Sholla]

Thanks for your help!

"The first thing I see is that eq. 5 is not correct. dΦ ≠ d(B⋅dA). "
Since Φ Ξ ∫B⋅dA, doesn't dΦ = d[∫B⋅dA] = d∫B(dA)cosθ?

"You missed something in eq. 6 WRT dA."
∫dAcosθ = A∫cosθ ? If so, is the integration variable dθ, and if so -- how does this come about?

"WRT eq 7 reconsider your integration limits."
Does the angle but the surface normal and the B-field not begin at 160° and end at 20°?

Thanks again for your help!

 

[Taulant Sholla]

Very helpful (and educational!). Thank you so much!

The flux integral should be taken over the entire area of the loop at a given instant in time. In particular, $\theta_i$ and $\theta_f$ are not bounds for the inner (flux) integral--in fact, they're bounds for the outer (time) integral, since $\theta$ is a function of time.

This actually simplifies things quite a bit, because as far as the inner integral is concerned, $\theta$ is a constant--meaning that factor of $\cos(\theta)$ can essentially be ignored when evaluating the integral. The expression $$ \int \cos(\theta) dA $$ then simply becomes $A \cos(\theta)$, where $A$ is the total area of the loop.

I actually like the way you've written the expression in 7 quite a bit, although I haven't seen it this way before. Using this notation, we have: $$ \frac{-NB}{R} \int_{\theta_i}^{\theta_f} d (A \cos(\theta)) = \left( \frac{-NB}{R} \right) A (\cos(\theta_f) - \cos(\theta_i)) $$ Note that this is identical to the expression given in the solution, except for your minor disagreement about the values of $\theta_i$ and $\theta_f$. I'll let you think about why that distinction actually doesn't matter in the end.

 

[Taulant Sholla]

Very helpful (and educational!). Thank you so much!

 

[Taulant Sholla]

One last question... in your last equation, how do you know to do d(cosθ)=Δcosθ rather than d(cosθ) = -sinθ ?

 

[VKint]

This is more or less notational. In the context of an integral, "d" doesn't really translate to "take a derivative"--think of it more as a label for the variable of integration. In particular, anytime you see an expression like this: $$ \int_a^b d[ \textrm{stuff} ] $$ the answer is always $$ \left. \textrm{stuff} \right]_a^b = \textrm{stuff}(b) - \textrm{stuff}(a). $$ For example, $\int_a^b dx = b - a$, despite the fact that "dx," interpreted as "the derivative of x," would just be a constant.

 

[gleem]

Since Φ Ξ ∫B⋅dA, doesn't dΦ = d[∫B⋅dA] = d∫B(dA)cosθ?

remember that in B⋅dA we are interested in the projected area in the direction of B, This projected area depends on the angel of B wrt dA and not dA itself. so
dΦ = ∫ dB⋅dA +∫B⋅d(dA ) = ∫_A∫_Θ B⋅dAd(cosΘ) = A∫_ΘsinΘdΘ which take care of eq 6 where you did not have an increment in the angle (dΘ) over which to integrate.

Does the angle but the surface normal and the B-field not begin at 160° and end at 20°?

Yes you are correct.

 

[Taulant Sholla]

remember that in B⋅dA we are interested in the projected area in the direction of B, This projected area depends on the angel of B wrt dA and not dA itself. so
dΦ = ∫ dB⋅dA +∫B⋅d(dA ) = ∫_A∫_Θ B⋅dAd(cosΘ) = A∫_ΘsinΘdΘ which take care of eq 6 where you did not have an increment in the angle (dΘ) over which to integrate.
Yes you are correct.

Thanks a bunch - this really gives me a lot of insight I did't really have before.