Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.48 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1840 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 5.19 V/m, (b) in the negative z direction and has a magnitude of 5.19 V/m, and (c) in the positive x direction and has a magnitude of 5.19 V/m

Answer 1

a) F = 15.6 10⁻¹⁹ k^ N, b) F = -1 10⁻¹⁹ k^ N,

c) F = (8.3 i^ + 73 k^) 10⁻¹⁹ N

Step-by-step explanation:

In this exercise we calculate the magnetic and electric force separately.

Let's start by calculating the magnetic force

F_m = q v x B

where bold letters indicate vectors, the modulus of this expression is

F_m = q v B

the direction is given by the right hand rule, where the thumb points in the direction of velocity, the fingers are extended in the direction of the magnetic field and the palm is in the directional ne force if the charge is positive

let's calculate the magnitude

F_m = 1.6 10⁻¹⁹ 1840 2.48 10⁻³

F_m = 7.3 10⁻¹⁹ N

we calculate the direction, the thumb is in the direction of + y,

fingers extended in the direction of -ax

the palm remains + z

therefore the magnetic force is in the direction of the positive side of the z axis

The electric force is

F_e = q E

give several possibilities for the electric field

a) electric field E = 5.19 V / m in the direction of + z

we calculate

F_e = 1.6 10⁻¹⁹ 5.19

F_e = 8.3 10⁻¹⁹ N

the total force is

F = F_m + F_e

F = (8.3 k ^ + 7.3 k ^) 10⁻¹⁹

F = 15.6 10⁻¹⁹ k^ N

b) in the -z direction

F_e = -8.3 10⁻¹⁹ k^ N

F = Fm + Fe

F = (-8.3 k ^ + 7.3 k ^) 10⁻¹⁹

F = -1 10⁻¹⁹ k^ N

c) in the + x direction

F_e = 8.3 10⁻¹⁹ i^ N

the total force is

F = (8.3 i^ + 73 k^) 10⁻¹⁹ N