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Figure $\mathrm{P} 20.3$ shows three edge views of a square loop with sides of length $\ell=0.250 \mathrm{m}$ in a magnetic field of magnitude 2.00 T. Calculate the magnetic flux through the loop oriented (a) perpendicular to the magnetic field, (b) $60.0^{\circ}$ from the magnetic field, and (c) parallel to the magnetic field.

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All right. So here we're dealing with sluts, which is five B. That's the ultimate a steel string. You temps area eight times. Co Sign of the angle between them. Had it sealed the scene of the loop. So in doesn't square loop. So they will just be all square elbow length of Luke stripes inside. Data for part A, you have data is equal to zero. Uh, so the flux will just be be tug's l squared. So it will be to Tesler times, uh, 0.25 meters quantity squared. So that's 18 That's 1 16 times 2/3. 18 our 0.1 to 5. Weber, this part D uh, you have angle. The angle is 19 minus 60 90 uh, minus 60 which is shown in the figure. So it's 30 degrees. Uh, so this implies that the flux is B. L squared Times co sign 30 and co sign 30. Aziz now is co signed. 30 is one over. It is one over square, so that will be equal to whoops. I meant to say that CO signed 30 is squared off to be over, too. Uh, since that's what, 3/2, so the flux is actually 125 year tons squared up to you to So it's equality 0.108 letters on Finally for part c, uh, perpendicular, uh, to this perpendicular field are to the saying of the loop, meaning that fights of B is a go to BL Square Co. Side 90 and that's zero's of the city code and that's it.