A group of particles is traveling in a magnetic field of unknown magnitude and d

Question

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.40 km/s in the +x-direction experiences a force of 2.06×1016 N in the +y-direction, and an electron moving at 4.40 km/s in the z-direction experiences a force of 8.10×1016 N in the +y-direction.

Part C

What is the magnitude of the magnetic force on an electron moving in the y-direction at 3.60 km/s ?

Part C

What is the magnitude of the magnetic force on an electron moving in the y-direction at 3.60 km/s ?

MasteringPhysics HWT-Google Chrome Secure httpsllsession.masteringphysics.com/ 78285326 Physics II for Engineering (01) Hwaz tem 2 Item 2 Correct A group of particles is traveling in a magnetic field of unknown magnitude and direction You observe that a proton moving at 1.40 km/sin the +z-direction experiences a Part C force of 2.06x10-16 Nin the tydrection, and an electron moving at 440 km/sin the -zdirection experiences a What is the magnitude of the magnetic force on an electron movingin the-ydirection at 360 km/s? force of 8.10x10 16N in the ty direction. Submit My Answers Give Up Part D What is the direction of this the magnetic force? in the zzplane) ° from the -r-direction Submit My Answers Give Ug e (9 a a S Help Close Resources T e previous l 2 of 10 l next, Continue 637 PM SW2017

Explanation / Answer

Let the magnetic field be vector-B = B1*i +B2*k and magnitude of charge on proton or electron be q
we have q*1400*iX(vector-B) = (2.06*10^-16)*j or
-1400*q*B2*j = 2.06*10^-16*j or
1400*q*B2 = -2.06*10^-16 ----------------------------------- 1
So B2 = (-2.06*10^-16)/(1400*1.6*10^-19) = -0.9196

Also similarly for negatively charged electron
we have -q*4400*kX(vector-B) = (8.10*10^-16)*j or
-4400*q*B1*j = 8.10*10^-16*j or
4400*q*B1 = -8.10*10^-16 -----------------------------------2 or
B1 = (-8.10*10^-16)/(4400*1.6*10^-19) = -1.15 or
B = sqrt[B1^2 +B2^2] = 1.47 T
angle with x axis = tan^-1[B2/B1] = 38degree or 52 degree with z axis in negative XZ plane

C vector force = -1.6*10^-19*3600(-j) X (B1*i +B2*k) =
= (-B1*k+B2*i)*1.6*10^-19*3600
Force will be in positive x and negative z plane making an angle 52 degrees with x axis and its magnitude F will be given by
F = 1.47*1.6*3.6*10^-16 N = 8.47*10^-16 N

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