2/25/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 1/2Master ItA proton moves perpendicular to a uniform magnetic field at a speed of 1.45 107 m/s and experiencesan acceleration of 1.55 1013 m/s2 in the positive x direction when its velocity is in the positive z direction.Determine the magnitude and direction of the field.Part 1 of 5 ConceptualizeThe magnetic field exerts a magnetic force on the proton, causing the acceleration. For a proton, speeds of and accelerations on the order of are quite reasonable.Part 2 of 5 CategorizeWe will use the particle under acceleration model to find the net force on the proton. This force is themagnetic force. The particle in a magnetic field model will give the magnitude and direction of the magneticfield.Part 3 of 5 AnalyzeBy Newton's second law, The magnetic force is the only force accelerating the proton, so wehave the following.We have a magnetic force in the direction ofPart 4 of 5 AnalyzeSolving for the magnitude of the magnetic field, we have the following.Part 5 of 5 AnalyzeBy the righthand rule, must be in the −y −y direction. This yields a magnetic force onthe proton in the +x +x direction when points in the direction.B107 m/s 1013 m/s2.F = ma = qvB sin 90°F = ma+xF = 1.67 × 10−27 kg 1.55 1.55 × 1013 m/s2 = 2.588 2.59 × 10−14 N.B = = = 1.1149 1.12 × 10−2 T = 11.14 11.2 mTFqv 2.59 2.59 × 10−14 N1.60 × 10−19 C 1.45 1.45 × 107 m/sBv +z2/25/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl
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