Question

Albert's laboratory is filled with a constant uniform magnetic field pointing straight up. Albert throws some charges into this magnetic field. He throws the charges in different directions, and observes the resulting magnetic forces on them. Given the sign of each charge and the direction of its velocity, determine the direction of the magnetic force (if any) acting on the charge.

Answer 1

Answer:

$\vec{F}=qB(v_y \hat{i} - v_x\hat{j})$

Explanation:

The force excerted by a magnetic field on a charged particle is given by the Lorentz force:

$\vec{F} = q \vec{v} \times \vec{B}$

Lets consider the z-direction of our coordinate system the same direction of the magnetic field, that is:

$\vec{B} = B \hat{k}$

Let us consider that the velocity of a given particle is:

$\vec{v} = v_x\hat{i} + v_y \hat{j} + v_z \hat{k}$

Therefore, since k×k = 0

$\vec{v} \times \vec{B} = (v_x\hat{i} + v_y \hat{j} + v_z \hat{k}) \times B\hat{k}\\\vec{v} \times \vec{B} = (Bv_x \hat{i} \times\hat{k}) + (Bv_y \hat{j}\times\hat{k})$

And since  i, j , k are a rigth hand system:

i × j = k

j × k = i

k × i = j  --> i × k= -j

$\vec{v} \times \vec{B} = Bv_x (-\hat{j}) + Bv_y \hat{i} = Bv_y \hat{i} - Bv_x\hat{j}$

Threfore, if the particle has charge q and velocity v = (vx,vy,vz), the magnetic force it will feel will be

$\vec{F}=qB(v_y \hat{i} - v_x\hat{j})$