The emf induced in the circuit is e b ℓ v 9 t 5 m 4

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The emf induced in the circuit is 10.0points A concave mirror has a radius of curvature of 0 . 726 m. An object is placed 2 . 41 m in front of the mirror. What is the position of the image? 011 10.0points In a thundercloud there may be an electric charge of 43 C near the top and 43 C near the bottom. These charges are separated by approximately 2 . 2 km. What is the magnitude of the electric force between them? The Coulomb constant is Since the charges are of opposite signs, this force is attractive. 012 10.0points A concave mirror has a radius of curvature of 0 . 726 m. An object is placed 2 . 41 m in front of the mirror. What is the position of the image?
8 . 98755 × 10 9 N · m 2 / C 2 . 013 10.0points
Version 036 – Final 58010 Fall 14 – yeazell – (58010) 7 Two parallel wires carry equal currents in the opposite directions. Point A is midway between the wires, and B is an equal distance on the other side of the wires. A B What is the ratio of the magnitude of the magnetic field at point A to that at point B ? 1. B A B B = 3 correct 2. B A B B = 1 2 3. B A B B = 1 3 4. B A B B = 0 5. B A B B = 2 3 6. B A B B = 5 2 7. B A B B = 4 8. B A B B = 10 3 9. B A B B = 4 3 10. B A B B = 2 Explanation: The magnetic field due to a long wire is B = μ 0 I 2 π r . Let the distance between the wires be r . The magnetic field at A due to the upgoing wire is B up,A = μ 0 I 2 π ( r/ 2) = μ 0 I π r . The right-hand rule tells us the direction is into the paper. Due to the fact that A is the same distance from both wires, the other wire gives a magnetic field at A of the same magnitude, also directed into the paper due to the right-hand rule. The total magnetic field at A is B A = 2 B up,A = 2 μ 0 I π r . Now, the field at B due to the upgoing wire is B up,B = μ 0 I 2 π (3 r/ 2) = μ 0 I 3 π r , again into the paper, while B down,B = μ 0 I 2 π ( r/ 2) = μ 0 I π r out of the paper. So B B = B down,B B up,B = μ 0 I π r parenleftbigg 1 1 3 parenrightbigg = 2 μ 0 I 3 π r , out of the paper. Comparing magnitudes, we find B A B B = 2 μ 0 I π r 2 μ 0 I 3 π r = 3 . 014 10.0points An electron is projected into a uniform mag- netic field given by vector B = B x ˆ ı + B y ˆ . The magnitude of the charge on an electron is e . x y z v electron B x B y B
Version 036 – Final 58010 Fall 14 – yeazell – (58010) 8 Find the magnetic force when the velocity of the electron is v ˆ . 6. eB y v ˆ i

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