How Do You Calculate Proton Velocity Components in a Magnetic Field?

[frankfjf]

Homework Statement

A proton moves through a uniform magnetic field given by B = (10i - 20j +30k)mT. At time t1, the proton has a velocity given by v = vxi + vyj + (2.0km/s)k and the magnetic force on the proton is F = (4.0 x 10^-17N)i + (2.0 x 10^-17N)k. At that instant, what are (a) vx and (b) vy?

Homework Equations

F = qv x B

The Attempt at a Solution

After taking the cross product I get an equation that looks like:

4 x 10^-17Ni + 2 x 10^-17Nj = e(.03vy + 40)i + (20 - .03vx)j - (.02vx - .01vy)k

It's at this point that I blank out. The Student Solutions Manual says that the next step is to equate corresponding components, but what do they mean by this? Set vy to zero then solve for vx and vice versa?

 

[frankfjf]

Nevermind! I figured it out.

 

[Proof.Beh]

we have F=|q|v*B then according to determinant (matrix.JPG) derive that
q(-40-3vy)i-q(20-3vx)j+q(10vy+20vx)k=F then we get

q(20-3vx)=0 --> vx=20/3
q(10vy+20vx)k=2*10^{-17} --> vy

Has it any problem?

Thanks.
Mr Beh

 

[bilal232lagcc]

What is the solution to this problem ... I am stuck at the same step as you were ? Please Help