hw10sol

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 – find all choices before making your selection. The due time is Central time. 001 (part 1 of 3) 10 points Given: μ 0 = 1 . 25664 × 10 - 6 T m / A . An infinite sheet of current, perpendicular to the y axis, located at y = 0 . The linear current density λ + x flows in the + x direction. By inspection we expect the magnetic field direction b B y , on the positive y side of the sheet (on the right-hand side of the sheet) to be in the + z direction, while the magnetic field on the negative y side of the sheet (on the left-hand side of the sheet) to be in the - z direction; y z x infinite 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 of length and w . B B w y z The “magneto motive force”, defined by M = I ~ B · d~s , is given by 1. none of 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 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 d s . L = Z 0 B ds - Z 0 B ds = B ‘ + B ‘ = 2 ‘ B . 002 (part 2 of 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 field 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 = 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 field of the Earth as in Figure. The rotation occurs in a region where the compo- nent of the Earth’s magnetic field 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 . r l dr v B in O What potential difference is developed be- tween its ends? Correct answer: 0 . 000246301 V. Explanation: Basic Concept: Motional emf E = B · l · v For a point on the bar, the velocity with which the point moves changes linearly with the distance from the point to the rotation center. So, the effective velocity for the whole bar equals: v eff = ω · l 2 = 2 πf · l 2 = 10 . 9956 m / s , and the induced emf in the bar is E = B · l · v eff = 0 . 000246301 V .
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