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

The force on a charged particle moving in a constant magnetic field can be calculated with the formula F = qνB sin θ, considering that the angle θ is the angle between the velocity and the magnetic field. Since the velocity here is perpendicular to the magnetic field, the sine of θ is 1. The right-hand rule is used to determine the direction of the force.

Step-by-step explanation:

Magnetic Force on a Charged Particle

To find the magnitude and direction of the force on a charged particle moving in a constant magnetic field, we can use the formula for the magnetic force:

F = qυB sin θ,

where q represents the electric charge, υ represents the particle's velocity, B represents the magnetic field strength, and θ is the angle between the velocity and the magnetic field vectors. In this question, the charge q is given as 1 (unit charge), the magnetic field B as 2ᴩ + 2ᴱ, and the velocity v as 22k. Since the velocity and magnetic field are perpendicular (velocity is along the z-axis and magnetic field is in the xy-plane), the sin θ is 1.

The force can be found using the cross product of the velocity vector and the magnetic field vector:

F = q(υ × B)

In terms of the direction, the right-hand rule can be used: if you point your fingers in the direction of the velocity and curl them towards the magnetic field direction, your thumb will point in the direction of the force exerted on a positive charge.

Answer 2

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)