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Calculating magnetic flux

by NewtonianAlch
Tags: flux, magnetic
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NewtonianAlch
#1
Sep7-11, 05:03 AM
P: 440
1. The problem statement, all variables and given/known data

A cube of edge length 0.05m is positioned as shown in the figure below. A uniform magnetic field given by B = (5 i + 3 j + 2 k) T exists throughout the region.



a) Calculate the flux through the shaded face.

2. Relevant equations

[itex]\phi[/itex] = B.A cos [itex]\theta[/itex]

3. The attempt at a solution

The area would simply be 0.0025m^2

I'm having trouble understanding how to get the angle and also how to interpret the given magnitude of the magnetic field, it's a vector quantity.

I thought at first the way to get the angle was to assume that the surface of the cube could be considered a vector as well, that way it would only have the j component since it's only got a direction in the y-axis.

Then using the formula for the angle between two vectors, I got 53.5 degrees, though I'm not too sure how to use the given magnetic field value.
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tiny-tim
#2
Sep7-11, 06:48 AM
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Hi NewtonianAlch!
Quote Quote by NewtonianAlch View Post
a) Calculate the flux through the shaded face.

I'm having trouble understanding how to get the angle and also how to interpret the given magnitude of the magnetic field, it's a vector quantity.
Forget angles, forget magnitude of the field
just do the inner product! (dot product)
the area can be represented by a vector of magnitude A in the normal direction, so just "dot" that with the field, and that's your flux!

(or you can "dot" it with the unit normal, and then multiply by the area same thing)
NewtonianAlch
#3
Sep7-11, 07:47 AM
P: 440
Hi tinytim,

Do you mean to say:

(5, 3, 2)[itex]^{T}[/itex].0.0025 which is (5*0.0025 + 3*0.0025 + 2*0.0025)

B.A

tiny-tim
#4
Sep7-11, 09:01 AM
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Calculating magnetic flux

No, (5,3,2).(the unit normal times 0.0025)

(btw, you can't write BT.A

it's either BTA or B.A )


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