(10%) Problem 12: A charge q = 1.152E-17 C moves with velocity v = ½ k = 5500 k
ID: 1779347 • Letter: #
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(10%) Problem 12: A charge q = 1.152E-17 C moves with velocity v = ½ k = 5500 k m/s in a uniform electric field E = Ex i = 75 i V/m and a uniform magnetic field B By j = 0.85 j T. Refer to the figure Randomized Variables q = 1.152E-17 C v = vk = 5500 k m/s ©theexpertta.com 17% Part (a) Express the magnitude of the electric force acting on the charge FE, in terms of Ex and q 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, F, acting on the charge in terms of By, vz and q 17% Part (e) Calculate the value of FB, in newtons ar rection agnetic torc e2 Positive y direction. Positive z direction. Negative z direction Negative x direction.Positive x direction.ONegative y direction Grade Summary Deductions Potential 0% 100% SubmissionsExplanation / Answer
(A) Fe = q E = q Ex
(B) Fe = (1.152 x 10^-17) (75 i)
Fe = 8.64 x 10^-16 N
(C) along +ve x axis
(d) Fb = q ( v x B)
v x B = v B (k x j) = v B (-i)
magnitude = q v B
(e) Fb = 5.39 x 10^-14 N
(f) along -ve x axis
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