D21 Ebert Just Mercy film review -1B XJust Mercy Seminar - Google Doc XMagnetic Field CW-C55720 HP X +webassign.net/web/Student/Assignment-Responses/last?dep=29325785#Q50A proton moves perpendicular to a uniform magnetic field B at a speed of 1.65 x 107 m/s and experiences an acceleration of 2.75 x 1013 m/s² in the positivex-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.Step 1By Newton's second law, F F = ma. The magnetic force is the only force accelerating the proton, so we have the following.F = ma = qvB sin 90°We have a magnetic force in the +x-direction ofF= = (1.67 × 10-27 kg)(2.75 ✔2.75 x 10¹3 m/s²) = 4.59254.59 x 10-14 N.Step 2Solving for the magnitude of the magnetic field, we have the following.4.59F✓x 10-14 NB =10-14 N)x 107 m/s)qv1.60 x 10-191.65x 10-2 T = 0.01738636364XYour response is off by a multiple of ten. mT8:25OSubmit Skip (you cannot come back)OSign outLOX⠀1 INTL

F = magnetic force = q( V × B) V = velocity vector B= magnetic field vector q = charge of…

A proton moves perpendicular to a uniform magnetic field B at a speed of 1.65 x 107 m/s and experiences an acceleration of 2.75 x 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.

A proton moves perpendicular to a uniform magnetic field B at a speed of 1.65 x 107 m/s and experiences an acceleration of 2.75 x 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.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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