Question

A particle with unit charge ​(qequals​1) enters a constant magnetic field Bequals2iplus2j with velocity vequals22k. Find the magnitude and direction of the force on the particle. Make a sketch of the magnetic​ field, the​ velocity, and the force.

Answer 1

Answer:

The question is poorly written, but il try to answer this in a generic way.

Remember that i will use only vectors here.

you say that B = (2, 2, 0) and v = (0, 0, 22)  and the charge is q = 1.

Where the units are missing in your question, but i guess that they are in the same system.

The magnetic force can be described as:

F = q*(vxB)

So we must solve the cross product, the generic way to write this is:

$\left[\begin{array}{ccc}i&j&k\\vx&vy&vz\\Bx&By&Bz\end{array}\right] = \left[\begin{array}{ccc}i&j&k\\0&0&22\\2&2&0\end{array}\right]$

now, we can calculate the determinant of this matrix, and we will get the solution of the cross product (vxB)

(vxB) = i*(-vz*By + vy*Bz) + j*(vz*Bx - vx*Bz) + k*(vx*By - vy*Bx)

(this generic equation you can use always, now, replace the values)

(vxB) = (-22*2)*i + (22*2)*j + 0*k = -44*i + 44*j

so the force is: q*(vxB) = q*(-44, 44, 0)