solution11_pdf

# To produce the indicated charge separa tion the

This preview shows pages 6–8. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
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 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern