Question

A group of particles is travelling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the +x direction experience a force of $\mathbf{2}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{16}}\mathit{N}$in the +y direction, and an electron moving at 4.75 km/s in the -z direction experience a force of $\mathbf{8}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{16}}\mathit{N}$in the +y direction (a) what are the magnitude and direction of the magnetic field? (b) what are the magnitude and direction of the magnetic force on an electron moving in the -y direction at 3.20 km/s.

Short answer

(a) The magnitude of the magnetic field is 1.46T and the direction of the magnetic field is $40°$.

(b) The magnitude of the magnetic force is$7.47\times {10}^{16}N$ and the direction of the magnetic force is $50°$.

Step-by-step solution

Definition of magnetic field

The term magnetic field is defined as the area around the magnet which behave like a magnet.

Determine magnitude and direction of the magnetic field

The given quantities are${\overrightarrow{v}}_d=(1.50\mathrm{km}/\mathrm{s})\widehat{i},{\overrightarrow{F}}_{\mathrm{p}}=\left(2.25\times {10}^{-16}\mathrm{N}\right)\widehat{\mathrm{j}},{\overrightarrow{\mathrm{v}}}_{\mathrm{e}}=-(4.75\mathrm{km}/\mathrm{s})\widehat{\mathrm{k}},{\overrightarrow{\mathrm{F}}}_{\mathrm{e}}=\left(8.85\times {10}^{-16}\mathrm{N}\right)\widehat{\mathrm{j}}$ proton with velocity and force experienced and electron with velocity and force experienced. Now the force with charge q, velocity v and magnetic field$\overrightarrow{B}$ on a particle is given

$\begin{array}{r}\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}\\ \begin{array}{rl}{F}_x\widehat{i}+F_y\widehat{j}+F_z\widehat{k}& =\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ v_x& v_y& v_z\\ B_x& B_y& B_z\end{array}\right|\\ & =q\left(v_y{B}_z-v_z{B}_y\right)\widehat{i}+q\left(v_z{B}_x-v_x{B}_z\right)\widehat{j}+q\left(v_x{B}_y-v_y{B}_x\right)\widehat{k}\end{array}\end{array}$

For positive charged particles $v_y=0,v_z=0,F_x=0,F_z=0$

So the force isrole="math" localid="1668320289486" $F_e=e{v}_e{B}_x$ and$B_x=\frac{F_e}{e{v}_e}$ now put all given values

$\begin{array}{r}{B}_z=\frac{-2.25\times {10}^{-16}\mathrm{N}}{\left(1.60\times {10}^{-19}\mathrm{C}\right)(1500\mathrm{m}/\mathrm{s})}\\ {\mathrm{B}}_z=-0.9375\mathrm{T}\end{array}$

And for the negative charged particles than the force is further now put all the values than

$\begin{array}{r}{B}_x=\frac{8.50\times {10}^{-18}\mathrm{N}}{\left(1.60\times {10}^{-19}\mathrm{C}\right)(4750\mathrm{m}/\mathrm{s})}\\ B_x=1.118\mathrm{T}\end{array}$

Now calculate magnitude and direction of magnetic

$\begin{array}{r}B=\sqrt{B_x^2+B_z^2}\\ B=\sqrt{(1.118T{)}^2+(-0.9375T{)}^2}\\ B=1.46T\\ \tan \left(\theta \right)=\frac{B_z}{B_x}\\ \tan \left(\theta \right)=\frac{-0.9375T}{1.118T}\\ \tan \left(\theta \right)=-0.8386\\ \theta ={\tan }^{-1}(-0.8386)\\ \theta ={40.0}^{\circ }\end{array}$

Hence the direction and magnitude of the magnetic field are$40.0°$ and 1.46T.

Determine the magnitude and direction of magnetic force

The velocity negative charged particle is${\overrightarrow{v}}_e-\left(3.20km/s\right)\hat{j}$ and the force is$\overrightarrow{\mathrm{F}}=q\overrightarrow{v}\times \overrightarrow{B}$ now put all the calculated and given values.

$\begin{array}{r}\overrightarrow{F}=q\overrightarrow{V}\times \overrightarrow{B}\\ \overrightarrow{F}=(-e)(3.2\mathrm{km}/\mathrm{s})\left(\widehat{-j}\right)\times \left(B_x\widehat{i}+B_z\widehat{k}\right)\\ \overrightarrow{F}=e\left(3.2\times {10}^3\mathrm{m}/\mathrm{s}\right)\left[B_x(-\widehat{k})+B_z\widehat{i}\right]\\ \overrightarrow{F}=\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(3.2\times {10}^3\mathrm{m}/\mathrm{s}\right)[-1.118\mathrm{T}(\widehat{\mathrm{k}})-0.9375\mathrm{T}(\widehat{\mathrm{i}}\left)\right]\\ \overrightarrow{F}=\left(-4.80\times {10}^{-18}\mathrm{N}\right)\widehat{i}+\left(-5.724\times {10}^{-18}\mathrm{N}\right)\widehat{k}\\ F_x=-4.80\times {10}^{-16}\mathrm{N},{\mathrm{F}}_{\mathrm{y}}=-5.724\times {10}^{-16}\mathrm{N}\end{array}$

Now calculate magnitude and direction of the magnetic force

$\begin{array}{r}F=\sqrt{F_x^2+F_y^2}\\ F=\sqrt{{\left(-4.80\times {10}^{-16}\mathrm{N}\right)}^2+{\left(-5.724\times {10}^{-16}\mathrm{N}\right)}^2}\\ \mathrm{F}=7.47\times {10}^{-16}\mathrm{N}\\ \tan \left(\theta \right)=\frac{F_z}{F_x}\\ \tan \left(\theta \right)=\frac{-5.724\times {10}^{-16}\mathrm{N}}{-4.80\times {10}^{-16}\mathrm{N}}\\ \tan \left(\theta \right)=1.1925\\ \theta ={\tan }^{-1}\left(1.1925\right)\\ \theta ={50.0}^{\circ }\end{array}$

Hence, the direction and magnitude of the magnetic force is$50.0°$ and $747x{10}^{-16}\mathrm{N}$.