Why use x for the B-field distance in MIT 8.02 problem 31-9

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They are integrating the B field over the area inside the loop. (Since I don't have the figure, I must go by the mathematics in their solution.) The loop must be distance, h, from the current carrying wire. The loop extends a distance, w, in the x direction and a distance L in the y direction.
 
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What do you mean "x direction"?

When I first did it I did this

[tex]\Phi = B \cdot A = \frac{\mu_0 I}{2\pi (h + w)} \cdot Lw[/tex]
 


flyingpig said:
What do you mean "x direction"?

When I first did it I did this

[tex]\Phi = B \cdot A = \frac{\mu_0 I}{2\pi (h + w)} \cdot Lw[/tex]

The magnetic field you used only applies at the very bottom of the loop. In general, B is given by mu_0*I/(2*pi*r), where r is the distance from the wire. You have to integrate over the area of the loop to get the flux.