Calculating emf in a Loop with Changing Magnetic Field

AI Thread Summary
The discussion focuses on calculating the electromotive force (emf) in a square loop of wire with a changing magnetic field described by B = 4.0 t^2y. The emf was initially calculated incorrectly, leading to confusion about the sign and direction of the result. Using Faraday's law, participants derived the correct flux and emf values, ultimately arriving at an emf of 4.8 x 10^-5 V, with clarification that the sign indicates direction rather than correctness. The importance of scientific notation and the proper application of Faraday's law were emphasized throughout the conversation. The final consensus is that the calculated emf is accurate, considering the directionality of the electric field.
123yt
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1. In the figure the square loop of wire has sides of length 2 cm. A magnetic field points out of the page; its magnitude is given by

B = 4.0 t^2y

where B is in teslas, t is in seconds and y is in meters. Determine the emf around the square at t = 1.50 s.


[URL]http://homework.phyast.pitt.edu/res/sc/gblanpied/courses/usclib/hrw8/hrwpictures/31-43.jpg[/URL]2. emf = -Change in Flux

The Attempt at a Solution



B = 4*t^2*y
Integrated the equation with respect to y over an interval of 0 to 0.02 meters (length of y), multiplied result by 0.02 (length of x). Final result was 0.000016*t^2, which should also be the flux.

Flux = 0.000016*t^2
Derived the equation, and got dFlux = 0.000032*t dt.
Inserted 1.5 into t and got 0.000048.

emf = -0.000048 V.
Answer isn't correct.
 
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123yt said:
1. In the figure the square loop of wire has sides of length 2 cm. A magnetic field points out of the page; its magnitude is given by

B = 4.0 t^2y

where B is in teslas, t is in seconds and y is in meters. Determine the emf around the square at t = 1.50 s.


[URL]http://homework.phyast.pitt.edu/res/sc/gblanpied/courses/usclib/hrw8/hrwpictures/31-43.jpg[/URL]2. emf = -Change in Flux
Start with Faraday's law:

emf = \int E\cdot ds = - \frac{d}{dt}\int B\cdot dA

dA = xdy

So the right side is:

-\frac{d}{dt}\int B\cdot dA = -\frac{d}{dt}\int 4xyt^2dy = -\frac{d}{dt}(t^2)\int 4xydy

Does that help?

AM
 
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Ok, and the integration of 4xydy from 0 to 0.02, and then multiplying that result by x is 0.000016.

emf = d/dt * t^2 * 0.000016.
Deriving gets 0.000032 * t
Plug 1.5 into t to get 0.000048 V.

Was the answer supposed to be positive?
 
123yt said:
Ok, and the integration of 4xydy from 0 to 0.02, and then multiplying that result by x is 0.000016.

emf = d/dt * t^2 * 0.000016.
Deriving gets 0.000032 * t
Plug 1.5 into t to get 0.000048 V.

Was the answer supposed to be positive?
I had missed the - sign in Faraday's law (since corrected). But that just determines the direction of the electric field/potential difference. The question does not tell you what direction to go around the loop, so the answer can be positive or negative. You have the correct answer as far as I can tell. Try using scientific notation: 4.8x10^-5 V and also try it with a - sign.

AM
 
Alright, it works now. Thanks for the help.
 
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