The radius of the circular conducting loop is R. Magnetic Field is decreasing at a constant rate \(\alpha\) (inwards to the plane of loop). Resistance per unit length of the loop is \(\rho\). The current in wire AB is? (AB is one of the diameters)

- mathslover

- katieb

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- mathslover

- mathslover

This is what I've tried :
\(i = \cfrac{1}{r} \cfrac{d\phi}{dt} = \cfrac{1}{r} \cfrac{S dB}{dt} \)
where \(\phi\) is magnetic flux. \(r\) is resistance. \(S\) is the area A of the circular loop.
I've been given with : \(\cfrac{r}{2 \pi l } = \rho \implies r = \rho \times 2 \pi l= \rho \times 2 \pi (2R)\)
and \(S = \pi (R^2)\)
So, we have :
\( i = \cfrac{1}{\rho \times 4 \pi R} \times \pi (R^2) \times (- \alpha) \)
\(i = \cfrac{R \alpha}{4 \rho } \)
And direction of current will be from B to A |dw:1433312077316:dw|

- mathslover

While the correct answer is "zero"
Can anyone help me with this?

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## More answers

- mathslover

Sorry, didn't get you.
We have \(\phi = \bf{B} . \bf{S} \)
Right? And the angle between S and B is 0 degrees. So, why would it be zero?

- mathslover

Ohk! Take your time buddy.

- mathslover

What I'm thinking is, we have to calculate current through AB.... so, that should make a difference, will it? @dan815

- mathslover

Yeah... and perpendicular to area vector is parallel to B vector.. ain't it?

- mathslover

Ouch! :P
The diameter has no surface, that's why..?

- mathslover

Oh, so, even if it was for the whole circular loop, the answer would have been zero..?

- dan815

starting over lol

- mathslover

uhmm...? :/ confused

- dan815

okay so, let me see if i got this question right, does it mean mag field is increasing over time, or its decreasing constnatly spatially as u get closed to the center

- abb0t

- mathslover

It simply means - \(\cfrac{dB}{dt} = - \alpha \)
Now if you differentiate it again, it will give you "zero" ... So, it is decreasing constantly !

- mathslover

Ignore that "-" before dB/dt

- dan815

are u sure, the way the wrote it can be interpreted both ways

- mathslover

"According to the Lenz's law emf's of the same magnitude in the clockwise direction are induced in the two loops into which the figure is divided. So, current is induced in the clockwise direction in the outer boundary but no current in wire AB"
This is what I got when I searched (just now) on google...
https://books.google.co.in/books?id=zlWRMWEcJG0C&pg=PA349&lpg=PA349&dq=The+radius+of+the+circular+conducting+loop+is+R.+Magnetic+Field+is+decreasing+at+a+constant+rate&source=bl&ots=saTfbsAdfw&sig=cETI6m4U9WxnEfIW1hiHEJlQLJU&hl=en&sa=X&ei=U6FuVeq0CtGUuATVmIKwBQ&ved=0CB4Q6AEwAA#v=onepage&q=The%20radius%20of%20the%20circular%20conducting%20loop%20is%20R.%20Magnetic%20Field%20is%20decreasing%20at%20a%20constant%20rate&f=false

- dan815

|dw:1433313628013:dw|

- dan815

yah there is a current induced in the outerloop

- dan815

but in the center wire i think the current tries to flow both ways and cancels out u have no flow

- mathslover

Why both ways? :/

- dan815

|dw:1433313980505:dw|

- dan815

it always wants ot go in counter clockwise or clockwise whatever result u got for the outerloop

- dan815

but if the top part wants to go clockwise and bottom loop too , and the total current is equal everywhere

- dan815

there u must have no flow in the middle

- mathslover

Oh WOW! Excellent. Thanks a lot Dan! :)

- dan815

sure thing :D

- dan815

make sure u understand why this argument works only for electromagnetic induced currents

- dan815

normally this wont happen in a circuit

- dan815

but since its EMF induced they wanna maintain that clockwise or ccw flow

- mathslover

Yeah, I understand that point. Thanks again! :)

- dan815

ok also u can show it mathematically

- dan815

if u calculate the 2 semicircle loops separetely

- dan815

|dw:1433314421990:dw|

- dan815

u can solve for each semi circle given that B field

- dan815

and ull end up with zero again

- dan815

cuz again ull get the current flow in opposite directions, which is more rigorous

- mathslover

I see! Thanks for making it clear to me from every angle. :) Mr. Lifesaver :D

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