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# The right hand rule tells us that the field resulting

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, the right hand rule tells us that the field resulting from this element is into the paper. Since this holds for every el- ement on the semicircle, the total field is also pointing into the paper. 003 10.0 points Calculate the magnitude of the magnetic field at a point 142 cm from a long, thin conductor carrying a current of 0 . 66 A. The permeability of free space is 1 . 25664 × 10 6 T · m / A. Correct answer: 9 . 29577 × 10 8 T. Explanation: Let : μ 0 = 1 . 25664 × 10 6 T · m / A , r = 142 cm = 1 . 42 m , and I = 0 . 66 A . The magnetic field of the wire is B = μ 0 I 2 π r = (1 . 25664 × 10 6 T · m / A) (0 . 66 A) 2 π (1 . 42 m) = 9 . 29577 × 10 8 T . 004 (part 1 of 2) 10.0 points A long, straight wire carries a current of 6 . 55 A. An electron travels at 82500 m / s parallel to the wire, 46 cm from the wire. The permeability of free space is 1 . 25664 × 10 6 N / A 2 and the charge on an electron is 1 . 6 × 10 19 C. What force does the magnetic field of the current exert on the moving electron? Correct answer: 3 . 75913 × 10 20 N. Explanation: Let : μ 0 = 1 . 25664 × 10 6 N / A 2 , v = 82500 m / s , q = 1 . 6 × 10 19 C , I = 6 . 55 A , and r = 46 cm = 0 . 46 m . Magnetic field produced by a straight, current-carrying wire is B = μ 0 I 2 π r = (4 π × 10 7 ) (6 . 55 A) 2 π (0 . 46 m) = 2 . 84783 × 10 6 T . The electron moves in the direction which is perpendicular to the direction of the magnetic field caused by the long wire. So the force on the electron in a magnetic field is F = B v q = (2 . 84783 × 10 6 T) (82500 m / s) × (1 . 6 × 10 19 C) = 3 . 75913 × 10 20 N . 005 (part 2 of 2) 10.0 points Which of the following statements is correct ? 1. The force on the electron is directed per- pendicular to the electron’s motion but with the limited information given we cannot de- termine if this force is directed towards or away from the wire. correct

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maini (nm7637) – hw10 – Shneidman – (12108) 3 2. The force on the electron is directed op- posite to the electron’s motion. 3. The force on the electron is perpendicular to the electron’s motion and perpendicular to the plane in which both the electron position and the wire lie, with not enough information given to determine the direction of the force. 4. The force on the electron is directed per- pendicular to the electron’s motion and must be directed towards the wire. 5. The force on the electron is directed per- pendicular to the electron’s motion and di- rected away from the wire. 6. The force on the electron is directed along the electron’s motion. 7. The force on the electron is perpendicular to the electron’s motion and perpendicular to the plane in which both the electron position and the wire lie, with the direction determined by the right hand rule. Explanation: As the electron moves, its current is oppo- site to the direction of its motion, and if this electron is in the same direction as the wire’s current, the force on the electron is directed perpendicular to the electron motion and to- wards the wire, and if the electron’s current is in the opposite direction as the wire’s current, the force is directed oppositely from the first situation. But the problem does not give the direction of the current and we do not know if the electron is repelled or attracted to the wire.
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