12homework10 - Husain, Zeena Homework 10 Due: Apr 5 2004,...

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Unformatted text preview: Husain, Zeena Homework 10 Due: Apr 5 2004, 4:00 am Inst: Sonia Paban 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 3) 10 points Given: = 1 . 25664 10- 6 Tm / A . An infinite sheet of current, perpendicular to the x axis, located at x = 0 . The linear current density + z flows in the + z direction. By inspection we expect the magnetic field direction b B x , on the positive x side of the sheet (on the right-hand side of the sheet) to be in the + y direction, while the magnetic field on the negative x side of the sheet (on the left-hand side of the sheet) to be in the- y direction; x y z infinite current sheet in y,z plane Current in + z direction, as shown by arrows. B B We choose the Amperian loop as the dashed line below with sides of length and w . B B w x y The magneto motive force, defined by M = I ~ B d~s, is given by 1. M = 2( + w ) B . 2. none of these. 3. M = 2 B . correct 4. M = ( + w ) B . 5. M = B . 6. M = 2 w B . 7. M = w B . Explanation: Basic concepts Amperes law I ~ B d~s = I . Solution The magneto motive force is given by M = I ~ B d~s, where the integral is around the Amperian loop. Only the sides of the loop contribute, since ~ B d~s = 0 on the top and bottom. On the sides of the loop, ~ B and d~s are parallel, so ~ B d~s = B ds . L = Z B ds- Z B ds = B + B = 2 B . 002 (part 2 of 3) 10 points The current encircled by the Amperian loop is 1. I enc = + z . correct 2. I enc = 2( + w ) + z . 3. I enc = w + z . 4. I enc = ( + w ) + z . 5. I enc = 2 + z . Husain, Zeena Homework 10 Due: Apr 5 2004, 4:00 am Inst: Sonia Paban 2 6. I enc = 2 w + z . Explanation: The current enclosed by the loop is given by the current per unit length times the length I enc = + z . 003 (part 3 of 3) 10 points Given : = 0 . 3 m , w = 0 . 065 m , and + z = 3 . 15 A / m . Find the magnitude of the magnetic field k ~ B k . Correct answer: 1 . 9792 10- 6 T. Explanation: From Amperes law L = I ~ B d~s = I enc 2 B = + z . Therefore B = 2 + z = 4 10- 7 2 (3 . 15 A / m) = 1 . 9792 10- 6 T . Note: Fourth of six versions. 004 (part 1 of 1) 10 points A metal bar spins at a constant rate in the magnetic field of the Earth as in Figure. The rotation occurs in a region where the compo- nent of the Earths magnetic field perpendic- ular to the plane of rotation is 2 . 9 10- 5 T. The bar is 0 . 54 m in length and its angular speed is 5 ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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12homework10 - Husain, Zeena Homework 10 Due: Apr 5 2004,...

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