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 2.49 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2450 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 3.37 V/m, (b) in the negative z direction and has a magnitude of 3.37 V/m, and (c) in the positive x direction and has a magnitude of 3.37 V/m?

Explanation / Answer

here,

velocity, v = 2450 j m/s

magnetic field, Bx = (- 2.49*10^-3 ) T

from Lortez equation, magnatic force to charge under influence of electric and magnetic field is given by expression

F = q*E + q*(v X B)

Where,
E is electric field
q is charge
v is velocity
B is magnetic field
X is cross b/e v and B

Case A:
Ez = 3.37 V/m

magnetic force, F = q ( E + (V X B) )

magnetic force, F = 1.6*10^-19 ( (3.37 k) + ( (0 i + 2450j + 0k) X (-2.49*10^-3 i + 0 j + 0 k ) ) )

magnetic force, F = 1.6*10^-19 ( (3.37 k) + 6.10 k) --------------- (1)

magnetic force, F = (1.52*10^-18) k N

magnitude, F = 1.52*10^-18 N

Case B :

Ez = - 3.37 V/m

from eqn 1, replacing value of Ez and solving :

magnetic force, F = 1.6*10^-19 ( (- 3.37 k) + 6.10 k)

magnetic force, F = (4.368*10^-19 k ) N

magnitude , F = 4.368*10^-19 N

Case C :

Ex = 3.37 V/m

from eqn 1, replacing value of Ez and solving :

magnetic force, F = 1.6*10^-19 ( (3.37 i) + 6.10 k)

magnetic force, F = (5.932*10^-19 i + 9.76*10^-19 k ) N

magnitude , F = sqrt((5.932*10^-19)^2 + (9.76*10^-19)^2 + 0^2 )

magnitude , F = 1.142*10^-18 N

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