Homework 4 Begin Date: 2/20/2018 2:00:00 PM -Due Date: 3/4/2018 11:59:00 PM End
ID: 1582605 • Letter: H
Question
Homework 4 Begin Date: 2/20/2018 2:00:00 PM -Due Date: 3/4/2018 11:59:00 PM End Date: 3/4/2018 11:59:00 PM (6%) Problem 7: A charge q = 5.28E-18 C moves with velocity v = vz k = 9500 k m/s in a uniform electric field E = 65 i V/m and a uniform magnetic field B = a, j = 0.85 j T. Refer to the figure. A V Randomized Variables q=5.28E-18 C va vzke 9500 k m/s Ctheexpertta.com 17% Part (a) Express the magnitude of the electric force acting on the charge FE, in terms of Ex and q. Grade Summary Deductioes Potential 0% 100% Submissions Attempts remaining: 12 (5% per attempt) tailed view 0 ACKSPACE CLEAR Hint Igive up! Submit Feedback: 2% deduction per feedback. Hints: 2% deduction per hint. Hints remaining:- 17% Part (b) Calculate the value of FE, in newtons. 17% Part (c) what is the direction of the electric force on the charge? 17% Part (d) Express the magnitude of the magnetic force, FB, acting on the charge in terms of By1% and q 17% Part (e) Calculate the value of FB, in newtons. 17% Part(f) what is the direction of the magnetic force on the charge?Explanation / Answer
a)
electric force by the electric field is given as
Fe = q Ex
b)
Fe = q Ex
inserting the values
Fe = (5.28 x 10-18) (65)i = 3.43 x 1016 i
hence magnitude of force = 3.43 x 1016 N
c)
direction of electric force : +X-direction
d)
magnetic force is given as
FB = q (vz x By)
e)
FB = q (vz x By)
inserting the values
FB = (5.28 x 10-18) ((9500 k) x (0.85 j))
FB = 4.3 x 10-14 (- i)
magnitude : 4.3 x 10-14 N
f)
negative X-direction
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