Calculating emf in a Loop with Changing Magnetic Field

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 15K views
123yt
Messages
30
Reaction score
0
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.
 
Last edited by a moderator:
Physics news on Phys.org
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:

[tex]emf = \int E\cdot ds = - \frac{d}{dt}\int B\cdot dA[/tex]

dA = xdy

So the right side is:

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

Does that help?

AM
 
Last edited by a moderator:
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.