How Do You Calculate Magnetic Flux Through a Cube's Face?

[NewtonianAlch]

Homework Statement

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.

Homework Equations

\phi = B.A cos \theta

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.

 

[tiny-tim]

Hi 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)

 

[NewtonianAlch]

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

 

[tiny-tim]

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

(btw, you can't write B^T.A …

it's either B^TA or B.A )