Calculate Charge Flow in Lenz's Law Induction Problem | Earth's Magnetic Field

[longk]

Homework Statement

At a certain place, Earth's magnetic field has magnitude B = 0.590 gauss and is inclined downward at an angle of 77.5° to the horizontal. A flat horizontal circular coil of wire with a radius of 10.0 cm has 1000 turns and a total resistance of 55.5 . It is connected in series to a meter with 140 resistance. The coil is flipped through a half-revolution about a diameter, so that it is again horizontal. How much charge flows through the meter during the flip?

Homework Equations

\phi=\int(B)(da)cos \theta
emf=-N(\frac{d\phi}{dt})
i=\frac{dq}{dt}

The Attempt at a Solution

now since i don't know how to approach the problem myself, i was thinking along the lines of using farady's law to find the change of the flux from the initial position to when end point and from there getting the emf and with the induced emf the current and from the current the amount of charges that flow through the wire, i was thinking of going that way but since the time is not given or any angular speed and such i have no other clue on how to approach the problem, i don't know if this counts as an attempt since I'm a bit of a noob here but if it doesn't ok
is this problem able to be solved without mention of time?

 

[gabbagabbahey]

You don't actually need to compute the EMF to find the total charge that flows through the loop (which is a good thing, because all you are able to calculate are the initial and final fluxes--- you have no idea how quickly the loop flips).

Instead, use Ohm's Law and Faraday's Law together to help you express dq/dt in terms of dphi/dt and the resistance in the loop. Then integrate both sides of your expression with respect to time, from sometime before the flip, to sometime after the flip...what does that give you?

 

[longk]

i was thinking along the same lines using the combination of ohms law and farady's law but i can never get the right answer, for work i tried this approach

V=IR EMF=-N\frac{d\phi}{dt}

IR=-N\frac{d\phi}{dt}

\frac{dq}{dt}R=-N\frac{d\phi}{dt}

\frac{dq}{dt}R=-N\frac{d(BAcos(\theta))}{dt}

\frac{dq}{dt}R=-NBA\frac{dcos(\theta))}{dt}

\frac{dq}{dt}R=NBAsin(\theta)\frac{d(\theta))}{dt}

change the dq/dt and d(\theta)/dt into delta and go from there, the delta t will cancel out and i cen get an expression for just delta q but every time i try my answer is off, i get 6.44e-6 when the correct is 1.85e-5 is my approach correct, oh yeah if my approach is in any way correct do i use degrees or radians because for this i use radians for delta \theta

 

[Dadface]

The charge is independent of the time so the dt on each side of the equation can be cancelled.When you integrate you should find that the charge is equal to the change of flux linkage divided by R.

 

[longk]

yeah now i see thanks guys, i was going the wrong way, i had to integrate not use delta