To produce the indicated charge separa tion the

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To produce the indicated charge separa- tion, the positive charges in the conductor experience upward magnetic forces while the negative charges in the conductor experience downward magnetic forces leaving the charge separation shown in the figure. Using the right-hand rule with vector F = qvectorv × vector B to produce this force on positive charges, the velocity vectorv must be directed from left to right ( ). F F v B B + + - - 012 10.0 points A copper bar has a constant velocity in the plane of the paper and perpendicular to a magnetic field pointed out of the plane of the paper. v B B If the bar is moving from left to right ( ), how are charges distributed on the bar? 1. The top will be negative and the bottom will be positive. correct 2. The top will be positive and the bottom will be negative. 3. Both the top and bottom of the bar will be positive. 4. Both the top and bottom of the bar will be negative. Explanation: Positive charges will move in the direction of the magnetic force, while negative charges move in the opposite direction. The right-hand rule with vector F = qvectorv × vector B pro- duces a force on positive charges such that the positive charges in the conductor experience downward magnetic forces while the negative charges in the conductor experience upward
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taherisefat (mt23852) – hw 11 – Opyrchal – (121012) 7 magnetic forces leaving the charge separation shown in the figure below. F F v B B - - + + The top will be negative and the bottom will be positive. 013 (part 1 of 3) 10.0 points In an AC electric generator, a rigid loop of wire rotates in an external magnetic field. Say the loop is positioned as shown at time t = 0. A S N waterfall sliding contacts Which graph best represents the induced current i ( t ) at later times? Take i > 0 for current flowing in direction shown by arrows. 1. i 0 vector t 2. i 0 vector t correct 3. i 0 vector t Explanation: The current is proportional to the time derivative of magnetic flux through the loop. The magnetic field is constant in this case but the loop area perpendicular to the magnetic field is varying because the loop is rotating. Therefore, the magnetic flux is proportional to the loop area which has the time depen- dence of cos( ωt ) in this case. In other words, the current should have the same time depen- dent form as the time derivative of cos ωt , i.e. , i ( t ) sin ωt . 014 (part 2 of 3) 10.0 points The AC generator consists of N = 9 turns of wire each of area A = 0 . 0873 m 2 and total resistance 11 . 7 Ω. The loop rotates in a mag- netic field B = 0 . 68 T at a constant frequency of 52 . 4 Hz. Find the maximum induced emf. Correct answer: 175 . 904 V. Explanation: Let : N = 9 turns , A = 0 . 0873 m 2 , B = 0 . 68 T , R = 11 . 7 Ω , and f = 52 . 4 Hz . Faraday’s Law for solenoid: E = - N · d Φ B dt . Ohm’s Law: I = V R First note that ω = 2 π f = 2 π (52 . 4 Hz ) = 329 . 239 rad / s .
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taherisefat (mt23852) – hw 11 – Opyrchal – (121012) 8 Using Faraday’s Law (Equation above) with the appropriate numerical values gives E max = N A B ω = (9 turns) (0 . 0873 m 2 ) × (0 . 68 T) (329 . 239 rad / s) = 175 . 904 V .
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