Question

 A 2 C charge travels with a velocity of ~v = (1.0ˆi+2.0ˆj+2.0 ˆk)m/s through a uniform magnetic field of (5.0ˆi + 4.0ˆj + 1.0 ˆk)T. Calculate the magnetic force exerted on the electron, in vector form.

Answer 1

Let’s write the magnetic force that acts on the particle travelling with a velocity in a magnetic field :

$$F=q(v\times B).$$

Let's first calculate the cross product $\vec{v}\times \vec{B}$:

$$\vec{v}\times \vec{B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & v_z \\ B_x & B_y & B_z \end{vmatrix},$$ $$\vec{v}\times \vec{B}=\hat{i}\begin{vmatrix} v_y & v_z \\ B_y & B_z \end{vmatrix}-\hat{j}\begin{vmatrix} v_x & v_z \\ B_x & B_z \end{vmatrix}+\hat{k}\begin{vmatrix} v_x & v_y \\ B_x & B_y \end{vmatrix},$$ $$\vec{v}\times \vec{B}=(v_yB_z-B_yv_z)\hat{i}-(v_xB_z-B_xv_z)\hat{j}+(v_xB_y-B_xv_y)\hat{k}.$$

Substituting components of $v$ and $B$, we get:

$$\vec{v}\times \vec{B}=(2-8)\dfrac{T\cdot m}{s}\hat{i}-(1-10)\dfrac{T\cdot m}{s}\hat{j}+(4-10)\dfrac{T\cdot m}{s}\hat{k},$$ $$\vec{v}\times \vec{B}=(-6\ \dfrac{T\cdot m}{s})\hat{i}+(9\ \dfrac{T\cdot m}{s})\hat{j}-(6\ \dfrac{T\cdot m}{s})\hat{k}.$$

Finally, we can find the magnetic force:

$$F=2\ C\cdot[(-6\ \dfrac{T\cdot m}{s})\hat{i}+(9\ \dfrac{T\cdot m}{s})\hat{j}-(6\ \dfrac{T\cdot m}{s})\hat{k}],$$ $$F=(-12\ N)\hat{i}+(18\ N)\hat{j}-(12\ N)\hat{k}.$$