Answer:
[tex]\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N[/tex]
Explanation:
The total force on the particle is given by
[tex]\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}[/tex]
Then, by replacing we have:
[tex]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[/tex]
where the cross product can be made with the determinant method.
Hope this helps!!
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)
