A proton travels through un form magnetic and electric fields. The magnetic fiel

Question

A proton travels through un form 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 ahd has a magnitude of 2490 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric fieid is (a) in the positive z direction and has a magnitude of 3.65 V/m, (b) in the negative z direction and has a magnitude of 3.65 V/m, and (e) in the positive x direction and has a magnitude of 3.65 V/m? UnitsT N 1.36e-18 umber 7.854e-19 9.7828

Explanation / Answer

here,

q = 1.6 * 10^-19 C

magentic feild , B = - 1.97 i mT = - 0.00197 i T

velocity of proton , v = 2490 j m/s

a)

electric feild , E = 3.65 k V/m  

the net force , F = q * ( v X B + E )

F = 1.6 * 10^-19 * ( 2490 j X ( - 0.00197 i) + 3.65 k)

F = 1.37 * 10^-18 k N

the magnitude of force is 1.37 * 10^-18 N

b)

electric feild , E = - 3.65 k V/m  

the net force , F = q * ( v X B + E )

F = 1.6 * 10^-19 * ( 2490 j X ( - 0.00197 i) - 3.65 k)

F = 2 * 10^-19 k N

the magnitude of force is 2 * 10^-19 N

c)

electric feild , E = 3.65 i V/m  

the net force , F = q * ( v X B + E )

F = 1.6 * 10^-19 * ( 2490 j X ( - 0.00197 i) + 3.65 i)

F = (7.84 k + 5.84 i) * 10^-19 N

the magnitude of force |F| = sqrt(7.84^2 + 5.84^2) * 10^-19

|F| = 9.78 * 10^-19 N

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