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 )

 

[rwisz]

To calculate the magnetic flux through the shaded face of the cube, we can use the formula for magnetic flux, which is given by φ = B·A·cosθ, where B is the magnetic field, A is the area of the face, and θ is the angle between the magnetic field and the normal vector of the face.

In this case, the magnetic field is given as B = (5 i + 3 j + 2 k) T, which means that the magnetic field has components in all three directions, x, y, and z. To find the magnitude of the magnetic field, we can use the formula |B| = √(Bx^2 + By^2 + Bz^2), where Bx, By, and Bz are the x, y, and z components of the magnetic field, respectively. In this case, the magnitude of the magnetic field is |B| = √(5^2 + 3^2 + 2^2) = √38 T.

Now, to find the angle θ, we need to find the dot product between the magnetic field and the normal vector of the shaded face. The normal vector of the shaded face is in the y direction, so it can be written as n = (0 j + 1 k). The dot product between B and n is given by B·n = (5 i + 3 j + 2 k)·(0 j + 1 k) = 3 j + 2 k.

Therefore, the angle between B and n can be found using the formula cosθ = (B·n)/(|B||n|) = (3 + 2)/(√38√2) = 0.707. Taking the inverse cosine, we get θ = 45 degrees.

Substituting these values into the formula for magnetic flux, we get φ = (√38 T)(0.0025 m^2)(cos45) = 0.042 Nm^2.

In conclusion, the magnetic flux through the shaded face of the cube is 0.042 Nm^2.