# 2 ev 1 6 10 19 j 1 kev 1000 ev 48 082 kev 008

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2 · eV 1 . 6 × 10 19 J × 1 keV 1000 eV = 48 . 082 keV . 009 10.0points Find the magnitude of the emf induced around the loop in the figure. The 0 . 638 T uniform magnetic field is directed into the plane of the circuit and the 30 . 6 cm long con- ductor moves at a speed of 3 . 14 m / s. 0 . 638 T 0 . 638 T 008 10.0points A thin, horizontal copper rod is 1 . 02819 m long and has a mass of 23 . 9052 g. The acceleration of gravity is 9 . 8 m / s 2 . What is the minimum current in the rod that will cause it to float in a horizontal mag- netic field of 2 . 03123 T? 16 . 3 Ω I 30 . 6 cm 3 . 14 m / s
Version 010 – Midterm 3 2PM Sp16 – yeazell – (55745) 6 An electron is in a uniform magnetic field B that is directed out of the plane of the page, as shown. v e B B B B When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed 1. toward the top of the page. 2. out of the page. 3. into the page. 4. toward the right 5. toward the left 6. toward the bottom of the page. correct Explanation: The force on the electron is vector F = q vectorv × vector B = - e vectorv × vector B. The direction of the force is thus hatwide F = - hatwide v × hatwide B , pointing toward the bottom of the page , us- ing right hand rule for hatwide v × hatwide B , and reversing the direction due to the negative charge on the electron.