Α cos 2 α 006 part 2 of 3 10 points what is the

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α cos 2 α 006 (part 2 of 3) 10 points What is the power lost to dissipation in the circuit? 1. P = 1 R ( B ‘ v sin α ) 2 2. P = B 2 2 v 2 R cos α sin 2 α 3. P = 1 R ( B ‘ v cos α ) 2 correct 4. P = B ‘ v R sin α 5. P = B 2 2 R cos 2 α 6. P = B ‘ v R sin 2 α Explanation: For current I and resistance R , the power dissipation is P = I 2 R . And from (4) of Part 1 P = I 2 R = 1 R B ‘ v 2 R = B 2 2 v 2 R = B 2 2 v 2 R cos 2 α .
Husain, Zeena – Homework 10 – Due: Apr 5 2004, 4:00 am – Inst: Sonia Paban 4 007 (part 3 of 3) 10 points (denote the original steady-state speed as v 1 ) Reverse the direction of B . The new steady-state speed v 2 is 009 (part 2 of 3) 10 points The direction of the induced magnetic field at the center of the circulating eddy current is Explanation: O
008 (part 1 of 3) 10 points A pendulum consists of a supporting rod and a metal plate (see figure). The rod is pivoted at O . The metal plate swings through a region of magnetic field. Consider the case where the pendulum is entering the magnetic field region from the left. O enter 010 (part 3 of 3) 10 points The direction of the force which the magnetic field exerts is 1. along the direction of swing. 2. opposite to the direction of swing. cor- rect 3. along the rod toward the pivot point. 4. out of the plane. 5. into the plane. F B i Explanation:
Husain, Zeena – Homework 10 – Due: Apr 5 2004, 4:00 am – Inst: Sonia Paban 5 Because the magnetic field only exerts a force on the current segment already in the magnetic field region, the net magnetic force is opposite to the direction of swing, see the figure in the explanation of the previous Part. 011 (part 1 of 1) 10 points A coiled telephone cord has 85 turns, a cross-
sectional diameter of 4 . 21 cm, and an un- stretched length of 32 cm. Determine an approximate value for the self-inductance of the unstretched cord.
012 (part 1 of 1) 10 points Given: The resistivity of copper is 1 . 7 × 10 - 8 Ω m .

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