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Unformatted text preview: homework 21 – KANUNGO, ANIL – Due: Oct 25 2007, 4:00 am 1 Question 1, chap 30, sect 5. part 1 of 2 10 points Consider the setup shown in the sketch. The conducting slab has a thickness b and a width a . The magnetic field is constant which points into the paper. For this part of the problem the voltmeter is disconnected from the circuit. There is a steady flow of a horizontal current I moving from left to right, or a steady flow of electrons from right to left with an average drift velocity v d . In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. L b a I P 1 P 2 V vector B vector B y x How much electric force is produced by the magnitude of the corresponding electric field? 1. E = 3 v d B 2. E = v d / (2 B ) 3. E = I B 4. E = v 2 d B 5. E = v d B correct 6. E = a v d b B 7. E = v d I B 8. E = a v d B Explanation: The cancellation of the magnetic force and electric force implies that q v d B = q E or E = v d B . Question 2, chap 30, sect 5. part 2 of 2 10 points A voltmeter (with an internal resistance less than infinity) is connected to the system, where the contact points are on the upper and lower surfaces and are in the same vertical plane. Choose the correct answer for the case where the sign of the charge of current carri- ers are either negative (electrons) or positive (holes). 1. The direction of the current through the voltmeter is downward for either positive or negative charge carriers. 2. The direction of the current through the voltmeter is downward for positive charge car- riers and upward for negative charge carri- ers. 3. The direction of the current through the voltmeter is upward for positive charge carri- ers and downward for negative charge carriers. correct 4. The direction of the current through the voltmeter is upward for either positive or neg- ative charge carriers. 5. The current through the voltmeter is zero for either positive or negative charge carri- ers. Explanation: First consider the negative charge carrier case. Applying the right hand rule, the force on electrons moving to the left is downward. This generates a counterclockwise electron flow, or a clockwise current. In other words at the voltmeter, the current flows downward. homework 21 – KANUNGO, ANIL – Due: Oct 25 2007, 4:00 am 2 For the positive charge carrier case, the force on the positive charged carrier is again downward. This generates a counterclockwise current. At the voltmeter now the current flows upward. Question 3, chap 30, sect 5. part 1 of 1 10 points A copper strip (8 . 47 × 10 22 electrons per cubic centimeter) 4 . 1 cm wide and 0 . 13 cm thick is used to measure the magnitudes of unknown magnetic fields that are perpendic- ular to the strip....
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This homework help was uploaded on 04/02/2008 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
- Fall '08