Question

1. An infinitely long filamentary wire carries a current of I A along Z-axis in the +z direction. Calculate the following: a) B at (x0, yo, z0) b) The flux through the square loop described by a spsb,cs:<d, $ = 90° (20 points)

Answer 1

ve 4G - H ) 11:35 AM 2- axe no=-3 z=offt y yo= 1 Zo=0 a ber 1- 1-1A point of intere H = B = I (cosoda - Cood) ad apps aixão B= A ( Cosoda- loso,) and & angle subtend by upper and to the pont of intorunt & outer anglee substended by conductor to the I P = √ x2 + y2 = Vio for infinite long conductor La zo di = 180 B = M1 X2 ad = Hlad UTP 20P _ î o

ve 4G - . - 11:35 AM 8 Borg ad = qdx ad O = 220* (-30-an) - - (6+10 ay + 2x10 an) _ î o

The equation for the magnetic flux through surface S has a form of 0 = / B cos ads The magnetic induction B can be expressed as the magnetic induction near the long straight conductor in a following form: where r is a distance between the conductor and the place, where we find the size of the magnetic induction. Because the distance r depends on the position in the square, we use it as integration unknown and express all other quantities in the magnetic flux equation using this unknown. Element of area ds will be expressed as ds = a d. where a is a side of the square parallel to the conductor and dr is an element part of the side perpendicular to the conductor, which should be expressed using integration unknown. Side of the blue triangle in the picture can be expressed using Pythagorean theorem as r = - di After derivation of this equation we get de = " " 22-" We install this form into the equation for the element of area ds ds= ard Now, we express the quantity cosa. Angle a is an angle between the direction of magnetic induction B and the perpendicular to the area. As we can see in the picture, the same angle is between the distance rand side of the square. We express cos ausing goniometric functions as cosa = 2 After installing for the r we get cosa = V2 - df COS Q = expressed using one unknown r. We can Now we have all parts of the equation for magnetic flux install all these parts into the equation = B cos ads Jd lar 12 d? Integration will be done over interval from d, to d, where the distance r differs. - Mold1 . -dr = auol d2 22 Ja 27 di

Return to call 05:34 O 4G ..l 0 11:34 AM d = a moz S for = ahol en dag 0 = za di 2A = IxAxxio7xl lny é = 2x107 lny d = 2x107 eny