hw10sol - Flores Michael Homework 10 Due Apr 5 2004 4:00 am...

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Flores, Michael – 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 – fnd all choices be±ore making your selection. The due time is Central time. 001 (part 1 o± 3) 10 points Given: μ 0 = 1 . 25664 × 10 - 6 T m / A . An infnite sheet o± current, perpendicular to the y axis, located at y = 0 . The linear current density λ + x ²ows in the + x direction. By inspection we expect the magnetic feld direction b B y , on the positive y side o± the sheet (on the right-hand side o± the sheet) to be in the + z direction, while the magnetic feld on the negative y side o± the sheet (on the le±t-hand side o± the sheet) to be in the - z direction; y z x infnite current sheet in z,x plane Current in + x direction, as shown by arrows. B B We choose the Amperian loop as the dashed line below with sides o± length and w . B B w y z The “magneto motive ±orce”, defned by M = I ~ B · d~s, is given by 1. none o± these. 2. M = ( + w ) B . 3. M = 2 ‘B . correct 4. M = 2 ( + w ) B . 5. M = w B . 6. M = ‘B . 7. M = 2 w B . Explanation: Basic concepts Ampere’s law I ~ B · d~s = 0 . Solution The magneto motive ±orce is given by M = I ~ B · d~s, where the integral is around the Amperian loop. Only the sides o± the loop contribute, since ~ B · d~s = 0 on the top and bottom. On the sides o± the loop, ~ B and d~s are parallel, so ~ B · d~s = B ds . L = Z 0 B ds - Z 0 B ds = B ‘ + B ‘ = 2 ‘B . 002 (part 2 o± 3) 10 points The current encircled by the Amperian loop is 1. I enc = ( + w ) λ + x . 2. I enc = ‘λ + x . correct 3. I enc = 2 w λ + x . 4. I enc = w λ + x . 5. I enc = 2 ( + w ) λ + x .
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Flores, Michael – Homework 10 – Due: Apr 5 2004, 4:00 am – Inst: Sonia Paban 2 6. I enc = 2 ‘λ + x . Explanation: The current enclosed by the loop is given by the current per unit length times the length I enc = ‘λ + x . 003 (part 3 of 3) 10 points Given : = 0 . 2 m , w = 0 . 084 m , and λ + x = 2 . 65 A / m . Find the magnitude of the magnetic ±eld k ~ B k . Correct answer: 1 . 66504 × 10 - 6 T. Explanation: From Ampere’s law L = I ~ B · d~s = μ 0 I enc 2 ‘B = μ 0 ‘λ + x . Therefore B = μ 0 2 λ + x = µ 4 π × 10 - 7 2 (2 . 65 A / m) = 1 . 66504 × 10 - 6 T . Note: Sixth of six versions. 004 (part 1 of 1) 10 points A metal bar spins at a constant rate in the magnetic ±eld of the Earth as in Figure. The rotation occurs in a region where the compo- nent of the Earth’s magnetic ±eld perpendic- ular to the plane of rotation is 1 . 6 × 10 - 5 T. The bar is 1 . 4 m in length and its angular speed is 5 π .
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This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.

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hw10sol - Flores Michael Homework 10 Due Apr 5 2004 4:00 am...

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