edigenwileyplus com/edugen/student/maintruni Introductory Physics (PHYs 1303, 13

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edigenwileyplus com/edugen/student/maintruni Introductory Physics (PHYs 1303, 1304, 1307, 130a) Halliday, Fundamentals of Physics, 10e Chapter 28 Problem 010 2Your answer is partially correct. Try again. A proton travels through uniform magnetic and electric feids The is in the direction and has a magnitude of 190 mr. At one instant the velocity or the preten is in the positve y directien aed has and had a magnitude or 4.61 VIm, and (c) in the magnitude of the net terce acting on the proton r the electric field is (a) n the positive z direction and has a magnitude of 4.61 vim, (b) in the negative direction positive direction and has a mi of+61 VIm (a) a (b) Number click you would like to show work for this question A Division of aabt.Nle.ASana Aa Rights Renervd

Explanation / Answer

For a particle moving in the +y and in a -x B field the force will be in the + z direction

magnitude = q*v*B = 1.60 x 10^-19 x 1990 x 1.90 x 10^-3 = 6.05 x 10^-19N

a)

the electric field creates a force = E*q in the +z direction

Fe = 4.61 x 1.60 x 10^-19

Fe = 7.376 x 10^-19N

So F = 6.05 x 10^-19 + 7.376 x 10^-19

F = 1.3426 x 10^-18N

b)

Now F = 6.05 x 10^-19 - 7.376 x 10^-19 = -1.326 x 10^-19 N

c)

Now add them as vectors so

F = sqrt((6.05 x 10^-19)^2 + (7.376 x 10^-19)^2) = 9.5396 x 10^-19 N

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