Question

A particle with a charge of -1.24x10^8 C is moving with instantaneous velocity (1.49x10^4m/s) î +(-3.85x10^4m/s) ĵ.​ What is the force exerted on this particle by a magnetic field

(a) (1.40T)î+ (-2.30T)k̂

(b) (-1.25T)î+ (2.00T)ĵ + (-3.50T)k̂

Answer 1

(a) 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 us first calculate the cross product $\vec{v}\times \vec{B}$:

$$\vec{v}\times \vec{B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & 0 \\ B_x & 0 & B_z \end{vmatrix},$$ $$\vec{v}\times \vec{B}=\hat{i}\begin{vmatrix} v_y & 0 \\ 0 & B_z \end{vmatrix}-\hat{j}\begin{vmatrix} v_x & 0 \\ B_x & B_z \end{vmatrix}+\hat{k}\begin{vmatrix} v_x & v_y \\ B_x & 0 \end{vmatrix},$$ $$\vec{v}\times \vec{B}=(v_yB_z)\hat{i}-(v_xB_z)\hat{j}+(-v_yB_x)\hat{k},$$

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

$$\vec{v}\times \vec{B}=(8.85\cdot10^4\ \dfrac{m\cdot T}{s})\hat{i}+(3.43\cdot10^4\ \dfrac{m\cdot T}{s})\hat{j}+(5.39\cdot10^4\ \dfrac{m\cdot T}{s})\hat{k}.$$

Finally, we can find the magnetic force:

$F=-1.24\cdot10^8\ C\cdot[(8.85\cdot10^4\ \dfrac{m\cdot T}{s})\hat{i}+(3.43\cdot10^4\ \dfrac{m\cdot T}{s})\hat{j}+(5.39\cdot10^4\ \dfrac{m\cdot T}{s})\hat{k}],$

$$F=(-1.1\cdot10^{13}\ N)\hat{i}-(4.25\cdot10^{12}\ N)\hat{j}-(6.68\cdot10^{12}\ N)\hat{k}.$$

(b) 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 us first calculate the cross product $\vec{v}\times \vec{B}$:

$$\vec{v}\times \vec{B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & 0 \\ B_x & B_y & B_z \end{vmatrix},$$ $$\vec{v}\times \vec{B}=\hat{i}\begin{vmatrix} v_y & 0 \\ 0 & B_z \end{vmatrix}-\hat{j}\begin{vmatrix} v_x & 0 \\ B_x & B_z \end{vmatrix}+\hat{k}\begin{vmatrix} v_x & v_y \\ B_x & B_y \end{vmatrix},$$ $$\vec{v}\times \vec{B}=(v_yB_z)\hat{i}-(v_xB_z)\hat{j}+(v_xB_y-v_yB_x)\hat{k},$$

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

$$\vec{v}\times \vec{B}=(1.35\cdot10^5\ \dfrac{m\cdot T}{s})\hat{i}+(5.21\cdot10^4\ \dfrac{m\cdot T}{s})\hat{j}-(1.83\cdot10^4\ \dfrac{m\cdot T}{s})\hat{k}.$$

Finally, we can find the magnetic force:

$F=-1.24\cdot10^8\ C\cdot[(1.35\cdot10^5\ \dfrac{m\cdot T}{s})\hat{i}+(5.21\cdot10^4\ \dfrac{m\cdot T}{s})\hat{j}-(1.83\cdot10^4\ \dfrac{m\cdot T}{s})\hat{k}],$

$$F=(-1.67\cdot10^{13}\ N)\hat{i}-(6.46\cdot10^{12}\ N)\hat{j}+(2.27\cdot10^{12}\ N)\hat{k}.$$