3. [418 Points] I DETAILS PREVIOUS ANSWERS MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER This question has several parts that must be completed sequentially. If you skip a part of the question, you Will not receive any points for the skipped part, and you Will not be able to come back to the skipped part. Tutorial Exercise a> 7 A proton moves perpendicular to a uniform magnetic field B at a speed of 2.35 x 10' mfs and experiences an acceleration of 1.55 x 1013 ml's2 in the positive x direction when its velocity is in the positive 2 direction. Determine the magnitude and direction of the field. Pan 1 OH 7 Conceptualize The magnetic field exerts a magnetic force on the proton, causing the acceleration. For a proton, speeds of 107 mfs and accelerations on the order of 10:13 mls2 are quite reasonable. Perl 2 of 5 - Categorize We will use the particle under acceleration model to find the net force on the proton. This force is the magnetic force. The particle in a magnetic field model will give the magnitude and direction of the magnetic field. Pan 3 of 5 iAnalyze By Newton's second law, 2F = ma. The magnetic force is the only force accelerating the proton, so we have the following. 2F: me = qu sin 90 We have a magnetic force in the +x cirection of F (1.67 x 10527 kg)( 1.55 y p 1.55 r 1013 mfszj = 2.59 v 5') 2.59 x 10'14N. Part 4 of 5 - Analyze Solving for the magnitude of the magnetic field, we have the following. N) F = l v . qr (mo . c)( e . mrs) 6.9 X Your response is off by a multiple of ten. 1' 10'2 T -2