What is the current induced in the loop when the area is π(b2 - a2)?

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

The area of the loop is π(b² - a²)

Emf induced E = Rate of change of flux = πk(b² - a²)

Resistance of the wire R = 2πλ(a+b)

Current in the loop = E/R = [k(b² - a²)]/[2λ(a+b)] .

This doesn't matches with the given answer .

 

[Charles Link]

Scratch that. I'll reply momentarily... (I was suggesting you factor $b^2-a^2=(b-a)(b+a)$, but from what I see what you have is incorrect)...[Editing note: The OP's solution is correct. The book has an error in their answer. It was difficult to see the OP's posting, so that in the following post, I misinterpreted the diagram. That later gets corrected in the discussion in the subsequent postings]. $\\$ In this problem, the EMF $\mathcal{E}$ is larger around the outside part of the loop at radius $b$ , than it is at radius $r=a$. The EMF will be a function of radius. For the outer part, the area in the $\mathcal{E}=- \frac{d \Phi}{dt}$ equation is $A=\pi b^2$. I think you may need to convert the resistance per unit length $\lambda$ to a resisitivity $\rho$, with the assumption that the loop has thickness $w$ which will ultimately drop out of the calculation, along with recognizing the other dimension is $l=b-a$, in order to compute resistivity $\rho$ from $\lambda$.$\\$ Meanwhile, current density $J=\sigma E$ with $\sigma=\frac{1}{\rho}$, where $E=E(r)$ is the induced electric field, and current density $J=J(r)$. The induced electric field at a radius $r$ is calculated as $E(r)= \frac{ \mathcal{E}(r)}{2 \pi r }$. The current in the loop is then the current density integrated over a cross section of area: $I=\int J(r) \, dA$, where $dA= w \, dr$, and $r$ goes from $a$ to $b$. $\\$ Perhaps there is a simpler approach, but this is how I would approach the solution. From what I can tell, this problem is somewhat unique, and is not a simple one. $\\$ And note: The EMF on a loop of radius $r$ is given by $\mathcal{E}(r)=-\frac{d \Phi}{dt}=- k \, \pi r^2$. You applied the EMF equation incorrectly in the OP. $\\$ And if my interpretation is correct, this is not an introductory E&M problem, but rather a somewhat advanced one. I provided more of the solution than is normally recommended for Homework Helpers to provide, because this one is not an introductory exercise. At least that is my interpretation. Again, perhaps someone else will see a simpler way of solving it. :) $\\$ And one other comment: The loop needs to be closed or you will get minimal current flowing. I question at this point the validity of the problem. If this is in an introductory course, the answer should presumably be zero. $\\$ Scratch my whole solution above: I thought the material covered the range between radius $r=a$ and $r=b$. My mistake. :-)

 

[PumpkinCougar95]

Is the length of loop $2 \pi (a+b)$ or something more ?

 

[Jahnavi]

Is the length of loop $2 \pi (a+b)$ or something more ?

Please neglect the extra length . What answer do you get ?

 

[PumpkinCougar95]

Same as yours.

 

[Jahnavi]

@Charles Link , do you now agree that my answer in OP is correct ?

 

[Charles Link]

@Charles Link , do you now agree that my answer in OP is correct ?

I think the length of the path may be $L=2 \pi (a+b)+2(b-a)$. $\\$ Otherwise, factor $b^2-a^2=(b-a)(b+a)$ and cancel the $b+a$ upstairs and downstairs.

 

[Jahnavi]

Thanks @PumpkinCougar95 and @Charles Link

 

[Jahnavi]

upstairs and downstairs.

 

[cnh1995]

Sanity check: You can see from the geometry that the current should be zero for b=a. Your answer in the OP gives you that while the book's answer doesn't.

 

[Charles Link]

Sanity check: You can see from the geometry that the current should be zero for b=a. Your answer in the OP gives you that while the book's answer doesn't.

I finally see the book's answer in the OP. (The OP is kind of difficult to read=I completely misinterpreted the diagram in my post 2, and the part with the book's answer isn't very legible). $\\$ Anyway, I see what the book did (incorrectly): They computed the EMF's as $\mathcal{E}_b=-k\, \pi b^2$ and $\mathcal{E}_a=-k \, \pi a^2$,and added them, and they didn't realize the inner EMF is in the opposite direction as the outer one in supplying EMF to the loop. The EMF's are both counterclockwise, but the loop reverses direction, so these two EMF's necessarily get opposite signs, which is the case if you simply assume the area of interest is $A=\pi (b^2-a^2)$, and have the path of integration simply progress continuously around the loop.