91影视

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

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

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

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

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

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

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

Now calculate magnitude and direction of magnetic

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

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

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

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

Now calculate magnitude and direction of the magnetic force

$\begin{array}{r}F=\sqrt{F_x^2+F_y^2}\\ F=\sqrt{{\left(鈭�4.80脳{10}^{鈭�16}\mathrm{N}\right)}^2+{\left(鈭�5.724脳{10}^{鈭�16}\mathrm{N}\right)}^2}\\ \mathrm{F}=7.47脳{10}^{鈭�16}\mathrm{N}\\ \tan 鈦�\left(胃\right)=\frac{F_z}{F_x}\\ \tan 鈦�\left(胃\right)=\frac{鈭�5.724脳{10}^{鈭�16}\mathrm{N}}{鈭�4.80脳{10}^{鈭�16}\mathrm{N}}\\ \tan 鈦�\left(胃\right)=1.1925\\ 胃={\tan }^{鈭�1}鈦�\left(1.1925\right)\\ 胃={50.0}^{鈭�}\end{array}$

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