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A cube of edge length $\ell=$ 2.5 $\mathrm{cm}$ is positioned as shown in Figure $\mathrm{P} 20.7$ . There is a uniform magnetic field throughout the region with components $B_{x}=+5.0 \mathrm{T}, B_{y}=$ +4.0 $\mathrm{T},$ and $B_{z}=+3.0 \mathrm{T}$ . (a) Calculate the flux through the shaded face of the cube. (b) What is the total flux emerging from the volume enclosed by the cube (i.e., the total flux through all six faces)?

a. \phi=.003125 T m^{2}

b. 0 T m^{2}

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Rutgers, The State University of New Jersey

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University of Winnipeg

All right. So here you have. Ah, Flux. Um uh, a cube. Excuse me. Um, with length, uh, side link. L right. And so you have to find the magnetic flux through this shaded region of the Q. Please excuse my terrible joining here, but basically have to find the flux. And and so flux, as we know, is equal to be a be times eight times co sign data on. So the magnetic field for the magnetic field were given. I had jihad on kay had X y z components. Um, and in this case, a CZ we know area will be L square. That's the, um, like the cheap square. Now, the field component that will that is perpendicular perpendicular to the cube face is the one very calculating. So for this cute face, the pot of the FISA B will be beasts of x times. L squared. Visa X is the X component right of the magnetic field of the magnetic field. So this will be five, uh, Tessa times L, which is 0.25 meters. That's 2.5 centimeters. It's quick. And this gives you a FISA, be of, um, Flux coming out of that phase of story 0.1 to 5. No times. 10 to the negative. Three Weber's That's and so that's part A For part B. V. Uh, You You want the net flux, right for part B you're looking at You're trying to find the net flux. Ah, through all six Q faces. So here you don't really have to calculate the flux. You're all cute since you note that cube has symmetry. So basically, if you find, um, what happens, uh, between this face and this face Ah, you can you can apply the same itude. You can say the symmetry holds for all faces where the top and bottom, the front and back. And nothing, no on the left and right. But so we already calculated for the right, it's 3.1 to 5 times 10 to the negative times, 10 to the negative. Three. Webber's, um four. Um, so let's call that 51 Okay, so if I won, is that fight too? Though component it's acting outwards will have Ah, negative boo. Right. So the component act s o the flux emerging out will be negative beasts of x times l squared. So that's negative 3.1 to 5 times 10 to the negative. That three Webber's s o fi one minus phi, too. So the net flocks, uh, through, uh, let's call that. Are you left? And right? Phase faces will be zero. Ah, and the same thing applies for top and bottom and front and back faces. So this implies that doves net flux overall is zero, and that's it.