An electron that has a velocity with x component 2.6 × 10 6 m/s and y component

Question

An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.8 × 106 m/s moves through a uniform magnetic field with x component 0.034 T and y component -0.19 T. (a)Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

Chapter 28, Problem 003 An electron that has a velocity with x component 2.6 x 106 m/s and y component 3.8 x 106 m/s moves through a uniform magnetic field with x component 0.034 T and y component -0.19 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity. (a) Number (b) Number Units So Units

Explanation / Answer

Given

electron of charge q = -1.6*10^-19 C

Velocity Vx = 2.6*10^6 m/s, magnetic field Bx = 0.034 T

and y components of

velocity Vy = 3.8*10^6 m/s , and magnetic field By = -0.19 T

we know that the magnetic force ona moving charge inside magnetic field is

F= q*Bv sin theta

here we can calculate the direction of both the magnetic field and velocoity

first magnetic field direction is Tan theta = By/Bx = -0.19/0.034

theta= -79.854 degrees

a)

magnitude is B = sqrt(Bx^2+By^2 +2Bx*By cos theta)

= sqrt(0.034^2+(-0.19)^2+2*0.034*(-0.19)cos 270) T

= 0.1930181 T

and the velocity vector direction is Tan theta = vy/vx = 3.8*10^6/2.6*10^6

theta= 55.61 degrees

magnitude is V = sqrtsqrt(vx^2+vy^2 +2vx*vy cos theta)

= sqrt((2.6*10^6)^2+(3.8*10^6)^2+2*2.6*10^6*(3.8*10^6)cos 90) T

= 4.604 *10^6 m/s

now the magnitude is F = q*B*V sin theta = -1.6*10^-19*0.1930181*4.604 *10^6*sin 135.464

F = (-9.9722381088768)*10^-14 N

b)

for proton the mangnetic force is F= q*BVsin theta, is F = (9.9722381088768)*10^-14 N

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