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Physics 23 Online
OpenStudy (anonymous):

(a) Evaluate the closed line integral of H about the rectangular path P1(2, 3, 4) to P2(4, 3, 4) to P3(4, 3, 1) to P4(2, 3, 1) to P1, given H = 3zax − 2x3az A/m. (b) Determine the quotient of the closed line integral and the area enclosed by the path as an approximation to (∇×H)y. (c) Determine (∇×H)y at the center of the area.

OpenStudy (anonymous):

So, what trouble are you having?

OpenStudy (anonymous):

i need help with b and c i completed part a

OpenStudy (anonymous):

Well, can you find the area enclosed by that loop?

OpenStudy (anonymous):

i do not know how to do it

OpenStudy (anonymous):

Draw the picture of those points.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

now what?

OpenStudy (anonymous):

So, what's the area of that shape?

OpenStudy (anonymous):

It should be a rectangle.

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Divide the value of the integral that you found by that area. That's the answer for part (b).

OpenStudy (anonymous):

so it would be 354/6?

OpenStudy (anonymous):

I don't know, I haven't done the integral - but that numerator seems large. Let me check.

OpenStudy (anonymous):

ok i know the correct answer for A is 354A

OpenStudy (anonymous):

The function is \[ \vec H = 3z\hat x - 2x^3 \hat z \] right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Okay, I agree. Yes, you are correct, the answer to be should be 354/6 = 59

OpenStudy (anonymous):

Now you must find the curl of the function \(\vec H \) and evaluate it at the center of the rectangle.

OpenStudy (anonymous):

how do i do that?

OpenStudy (anonymous):

If you don't know what a curl is, you should consult your book or Wikipedia. it is one of the three major vector differential operators.

OpenStudy (anonymous):

Wow Good job @Jemurray3 you couldent explained that better!

OpenStudy (anonymous):

*Could not

OpenStudy (anonymous):

so i should look up curl of center of the area?

OpenStudy (anonymous):

The definition of curl is the following: if \( \vec H = H_x \hat x + H_y \hat y + H_z \hat z \) then \[ \nabla \times \vec H = \left( \frac{\partial}{\partial y} H_z - \frac{\partial}{\partial z} H_y\right)\hat x + \left( \frac{\partial}{\partial z} H_x - \frac{\partial}{\partial x} H_z\right)\hat y + \left( \frac{\partial}{\partial x} H_y - \frac{\partial}{\partial y} H_x\right)\hat z\] So you need to calculate this function, and then plug in the coordinates of the center of the rectangle. Also, thank you Holly, but I didn't really explain anything :)

OpenStudy (anonymous):

thank you

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