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g A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 1.97 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1680 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 4.34 V/m, (b) in the negative z direction and has a magnitude of 4.34 V/m, and (c) in the positive x direction and has a magnitude of 4.34 V/m

Respuesta :

Answer:

a) 1.22*10^-18 N in the positive z direction

b) 1.65*10^-19 N in the negative z direction

c) (6.94*10^-19 N) in the positive x direction + (5.30*10^-19 N) in the positive z direction

Explanation:

See attachment for calculations

Ver imagen barackodam

(a) The electromagnetic force on the proton is 1.224 × 10⁻¹⁸ [tex]\hat k[/tex] N

(b) The force is 1.65 × 10⁻¹⁹ [tex](-\hat k)[/tex] N

(c) The force is (6.94 [tex]\hat i[/tex] + 5.29 [tex]\hat k[/tex]) × 10⁻¹⁹ N

Electromagnetic force on the proton:

Given a proton moving in the positive y-direction with a speed of :

v = 1680 m/s [tex]\hat j[/tex]

The magnetic field is in the negative x-direction with magnitude:

B = 1.97 mT [tex](-\hat i)[/tex]

(a) Electric field applied in positive z-direction :

E = 4.34 V/m [tex]\hat k[/tex]

The net force on the proton is iven by:

F = q (E + v×B)

where q is the charge on proton, given by:

q = 1.6×10⁻¹⁹ C

So,

F = 1.6×10⁻¹⁹( 4.34 [tex]\hat k[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )

F = 1.6×10⁻¹⁹ ( 4.34 [tex]\hat k[/tex] + 3.309 [tex]\hat k[/tex])

F = 1.224 × 10⁻¹⁸ [tex]\hat k[/tex] N

(b) Electric field applied in negative z-direction :

E = 4.34 V/m [tex](-\hat k)[/tex]

The net force on the proton is iven by:

F = q (E + v×B)

where q is the charge on proton, given by:

q = 1.6×10⁻¹⁹ C

So,

F = 1.6×10⁻¹⁹( 4.34 [tex](-\hat k)[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )

F = 1.6×10⁻¹⁹ ( 4.34 [tex](-\hat k)[/tex] + 3.309 [tex]\hat k[/tex])

F = 1.65 × 10⁻¹⁹ [tex](-\hat k)[/tex] N

(c) Electric field applied in positive x-direction :

E = 4.34 V/m [tex]\hat i[/tex]

The net force on the proton is iven by:

F = q (E + v×B)

where q is the charge on proton, given by:

q = 1.6×10⁻¹⁹ C

So,

F = 1.6×10⁻¹⁹( 4.34 [tex]\hat i[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )

F = 1.6×10⁻¹⁹ ( 4.34 [tex]\hat i[/tex] + 3.309 [tex]\hat k[/tex])

F = (6.94 [tex]\hat i[/tex] + 5.29 [tex]\hat k[/tex]) × 10⁻¹⁹ N

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