Question

A particle with positive charge q = 9.61 10-19 C moves with a velocity v = (5î + 4ĵ − k) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving particle, taking B = (4î + 3ĵ + k) T and E = (5î − ĵ − 2k) V/m. (Give your answers in N for each component.)

Answer 1

Answer: F = (1.15*10^-17i N - 9.61*10^-18j N - 2.88*10^-18k N)

Explanation:

Given

Charge of particle, q = 9.61*10^-19 C

Velocity of particle, v = (5i + 4j -k) m/s

Magnitude of magnetic field, B = (4i + 3j + k) T

E = (5i - j - 2k) V/m

Force, F = ?

The formula for this is

F = Eq + qv X B

F = q(E + v X B)

v X B = (5i + 4j - k) X (4i + 3j + k), using cross product, we have

v X B = (7i - 9j - k), adding E to it, we have

E + v X B = (12i - 10j - 3k), remember,

F = q(E + V X B), on multiplying by q, we get

F = (1.15*10^-17i N - 9.61*10^-18j N - 2.88*10^-18k N)

Answer 2

Answer:

$\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N$

Explanation:

The total force on the particle is given by

$\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}$

Then, by replacing we have:

$q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N$

where the cross product can be made with the determinant method.

Hope this helps!!