# Calculating magnetic flux

by NewtonianAlch
Tags: flux, magnetic
 P: 448 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 $\phi$ = B.A cos $\theta$ 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.
HW Helper
Thanks
P: 26,148
Hi NewtonianAlch!
 Quote by NewtonianAlch 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)
 P: 448 Hi tinytim, Do you mean to say: (5, 3, 2)$^{T}$.0.0025 which is (5*0.0025 + 3*0.0025 + 2*0.0025) B.A
 Sci Advisor HW Helper Thanks P: 26,148 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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