Calculate Force on Particle in Uniform Magnetic Field (2024)

Homework Statement

2 A particle with charge 94.5 nC is moving in a region where there is a uniform magnetic field of 0.450 T in the positive x-direction. At a particular instant of time, the velocity of the particle has components:
vx = - 1.68 x 10⁴ ms-1, vy = - 3.11 x 10⁴ ms-1 and vz = 5.85 x 10⁴ ms-1.
What are the components of the force on the particle at this time?

Homework Equations

F=qvBsin(theta)

The Attempt at a Solution

1st how do i get Greek letters into a post?
F_x=qvBsin0 = 0N
F_y=94.5*10^-9*-3.11*10⁴*.45sin90 = -1.322*10^-3N
F_z=94.5*10^-9*5.85*10⁴*.45sin90 = 2.49*10^-3N

The reason I am posting this is because it seemed way to easy so I think I may have totally missed/screwed something up?
Thanks for any help.

It is best to start from
F = q v x B
F = q(v_xi + v_yj + v_zk) x Bi = q B(v_x ixi + v_y jxi + v_z kxi)

calculate the three cross products separately and see what you get.

For special symbols, save the items below somewhere, then cut and paste from there.

α β γ δ ε θ λ μ ν π ρ σ τ η φ χ ψ ω Γ Δ Θ Λ Π Σ Φ Ψ Ω
∂ ∏ ∑ ← → ↓ ↑ ↔ ⇐⇑⇒⇓⇔
± − ÷ √ ∫ ½ ∞∴ ~ ≈ ≠ ≡ ≤ ≥ ° ∇∝

There are three cross products in Kuruman's last expression (i x i, j x i, and k x i)

i x i is not equal to i^2 (i is a vector)

pat666 said:

Hey Kuruman, thanks for the reply.
I chucked your equation in my 89 an came up with -7.144*10^-4i²-1.3225*10^-3ij+2.4877*10^-3ik
the only inconsistent value is the 1st 1 which i hope is simply because of the lack of sin(theta) since the force in the i plane would be zero?
also unsure of what "calculate the three cross products separately and see what you get." means? I can't see how to do three of them, is what I did what you meant for me to do?
Thanks

Listen to JaWiB. These cross products are vectors and your 89 appears not to be able to handle them correctly. You need to simplify cross products of unit vectors like ixj and write them as single unit vectors. Look up "cross product" and do it correctly.

pat666 said:

{0,2.48771E-3xz,1.32253E-3xy}

How do you interpret this? What are the three elements separated by commas and what do xz and xy stand for?

kuruman said:

Figure this out without using your 89 because it seems you do not understand what it's spitting back at you.

Downside of having a halfway decent calculator is that I forget the process of solving things like this almost immediately after the exam!

Reading the PF "cross product page" i get ixj=ijsin90=ij

I also found that "The cross product of two vectors A and B is a third vector (strictly, a pseudovector or axial vector) perpendicular to both of the original vectors, with magnitude equal to the product of their magnitudes times the (positive) sine of the angle between them, and in the direction determined by the right-hand rule."
This leaves me more confused about what my xy and xz's are??

pat666 said:

I also found that "The cross product of two vectors A and B is a third vector (strictly, a pseudovector or axial vector) perpendicular to both of the original vectors, with magnitude equal to the product of their magnitudes times the (positive) sine of the angle between them,

So the cross product is a vector. To specify it, you need to specify its magnitude and direction. First the magnitude. What is the product of the magnitude of i times the magnitude of j times sin90^o? Don't forget that i and j are unit vectors.

and in the direction determined by the right-hand rule."

Now for the direction. What does the right hand rule say about it?

pat666 said:

again trying to remember 11c, a unit vector is just a vector of length 1unit? so the magnitude of ixj must be 1?

Yes and yes. Now for the direction. It must be perpendicular to the xy plane, but there are two perpendiculars. By definition the +z direction (unit vector k) is in the ixj direction. So the cross product between two unit vectors is another unit vector in a direction perpendicular to the two (or zero). Back to my first post. Find the following cross products

ixi = ________

jxi = _______

kxi = ________

i found the cross product of v x B at once,

the vector i found perpendicular to both v and B is 0i -26585j - 13995k
Thats before multiplying through the charge, so after i do that it SHOULD give the magnitude of the force in component form.

does that sound right?

I used determinants to find cross product, all by hand.

i got F= 0i - 0.025j - 0.013k ??

got anything like this pat?