Problem 29.2 1 . 2 AEd ? BA = nT . 3 Submit Request Answer 1.0 mm 1.0 mm Part F . 5 . 4 Indicate the direction of vector B. .Problem 29.26 - Enhanced - with Expanded Hints An electron moves in the magnetic field B = 0.20i T with a speed of 1.0 x 10 m/s in the directions shown in (Figure 1) and (Figure 2). Part B For help with math skills, you may want to review: The Vector Cross Product What is magnetic force F on the electron for (Figure 2)? Figure Enter components of the force in piconewtons separated by commas. View Available Hint(s) VO AEd ? X Fa, Fy, F2 = PN Submit Request AnswerRevi What is the direction of B? Give your answer as an angle measured cow from the +r-axis. A proton moving in a uniform magnetic field with Express your answer in degrees. v1 = 1.03 x105 i m/s experiences force View Available Hint(s) F1 = 1.65 x10-16 / N. A second proton with 12 = 2.29 x106 j m/s experiences Hint 1. How to approach the problem F2 = - 4.07 x10-16 k N in the same field. Use the expression of the force exerted by a magnetic field on a moving charge and apply the right-hand rule to find the components of the field. Recall how to find the direction of a vector if you know its components. VO AEd ? 8 = Submit Request AnswerProblem 29.8 Part A The element niobium, which is a metal, is a superconductor (i.e., no electrical resistance) at temperatures below 9 K. However, the What is the maximum current in a straight, 3.80-mm-diameter superconducting niobium wire? superconductivity is destroyed if the magnetic field Express your answer with the appropriate units. at the surface of the metal reaches or exceeds 0.10 T. O ? Value Units Submit Request Answer Provide FeedbackProblem 29.2