# 007 100 points assume the mobile charge carriers are

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007 10.0 points Assume: The mobile charge carriers are either electrons or holes. The holes have the same magnitude of charge as the electrons. Note: In the figure, the point at the upper edge P 1 and at the lower edge P 2 have the same x coordinate. A constant magnetic field points out of the paper. There is a steady flow of a horizontal current flowing from left to right in the x direction. L b a I P 1 P 2 V vector B vector B y x 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 upward for positive charge carri- ers and downward for negative charge carriers. correct 3. The current through the voltmeter is zero for either positive or negative charge carri- ers. 4. The direction of the current through the voltmeter is upward for either positive or neg- ative charge carriers. 5. The direction of the current through the voltmeter is downward for positive charge car- riers and upward for negative charge carri- ers. Explanation: Assume that the charge carries are elec- trons.

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Version 003 – Midterm3 – li – (59050) 5 v d × B F = - e v d × B B v d The magnetic force on a negative charge points downwards. Hence an excess of positive charge develops along the upper edge. The induced electric field then points downwards, so P 1 is at a higher potential relative to P 2 . F + + + + + - - - - - The Hall current flows in the direction of the induced field, so it flows through the volt- meter from top to bottom. 008 10.0 points Two species of singly charged positive ions of masses 1 . 01 × 10 26 kg and 2 . 11 × 10 26 kg enter a magnetic field at the same location with a speed of 1 . 4 × 10 5 m / s. The charge on an electron is q e . If the strength of the field is 0 . 141 T and the ions move perpendicularly to the field find their distance of separation after they complete one half of their circular path. 1. 13.6354 2. 19.1681 3. 10.7999 4. 6.10348 5. 2.35043 6. 2.86073 7. 12.969 8. 8.11375 9. 8.83586 10. 7.28859 Correct answer: 13 . 6354 cm. Explanation: Let : q = 1 . 602 × 10 19 C , m 1 = 1 . 01 × 10 26 kg , m 2 = 2 . 11 × 10 26 kg , v = v 1 = v 2 = 1 . 4 × 10 5 m / s , and B = 0 . 141 T . The radius of the orbit as a function of the mass m of the particle is given by R = m v q B . The separation after one-half of the circular path is d = 2 ( R 2 R 1 ) = 2 ( m 2 v 2 m 1 v 1 ) q B = 2 ( m 2 m 1 ) v q B = 2 (2 . 11 × 10 26 kg 1 . 01 × 10 26 kg) 1 . 602 × 10 19 C × parenleftbigg 1 . 4 × 10 5 m / s 0 . 141 T parenrightbigg parenleftbigg 100 cm 1 m parenrightbigg = 13 . 6354 cm . 009 10.0 points A positively charged particle moving at 45 angles to both the z -axis and x -axis enters a magnetic field (pointing out of of the page), as shown in the figure below.
• Spring '08
• Turner
• Physics, Magnetic Field, Electric charge, r Biot-Savart Law

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