A proton travels through uniform magnetic and electric fields. The magnetic fiel

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.18 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1570 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.47 V/m, (b) in the negative z direction and has a magnitude of 5.47 V/m, and (c) in the positive x direction and has a magnitude of 5.47 V/m? Number _______ Units Number _______ Units Number _______ Units

Explanation / Answer

total force on a charge=q*(cross product of v and B)+q*E

let i and j are unit vectors along +ve x and +ve y axis.

given q=1.6*10^(-19) C

v=1570 j m/s

B=-2.18*10^(-3) i T

then cross product of v and B=3.4226 k

part a:

given E=5.47 k

then total force=q*(3.4226+5.47) k

=1.4228*10^(-18) k

force magnitude is 1.4228*10^(-18) N

force is in +ve z direction

force unit is newton.

part b:

given E=-5.47 k

then total force =q*(3.4226-5.47) k

=-3.2758*10^(-19) k

force magnitude=3.2758*10^(-19) N

direction=-ve z axis

unit= newton

part c:

given E=5.47 i

then total force=q*(3.4226 k +5.47 i)

force magnitude=q*sqrt(3.4226^2+5.47^2)=1.0324*10^(-18) N

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