91Ó°ÊÓ

$\stackrel{→}{\mathrm{B}}=(10\hat{\mathrm{i}}+20\hat{\mathrm{j}}+30\hat{\mathrm{k}})\mathrm{mT}$

$\stackrel{→}{\mathrm{v}}={\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}+(2.00\mathrm{km}/\mathrm{s})\hat{\mathrm{k}}$

$\stackrel{→}{{\mathrm{F}}_{\mathrm{B}}}=(4.0×{10}^{-17}\mathrm{N})\hat{\mathrm{i}}+(2.0×{10}^{-17}\mathrm{N})\hat{\mathrm{j}}$

Findthe values of ${\mathit{v}}_{\mathbf{x}}$and ${\mathit{v}}_{\mathbf{y}}$for the proton using the formula for magnetic force in terms of charge, magnetic field strength, and velocity of the proton.

Formulae are as follow:

$\stackrel{→}{F_B}=e\stackrel{→}{v}×\stackrel{→}{B}$

Where, F_B is magnetic force, v is velocity, m is mass, K.E. is kinetic energy, B is magnetic field, e is charge of particle.

The magnetic force experienced bytheproton is,

$\stackrel{→}{F_B}=e\stackrel{→}{v}×\stackrel{→}{B}$

role="math" localid="1663230978636" $(4.0×{10}^{-17})\hat{\mathrm{i}}+(2.0×{10}^{-17})\hat{\mathrm{j}}=1.6×{10}^{-19}\left[\left({\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}+(2.00\mathrm{km}/\mathrm{s})\hat{\mathrm{k}}\right)×\left((10\hat{\mathrm{i}}+20\hat{\mathrm{j}}+30\hat{\mathrm{k}}\right)\right]$$\stackrel{→}{v}×\stackrel{→}{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ v_x& v_y& 2.0\\ 10& -20& 30\end{array}\right|$

$\stackrel{→}{\mathrm{v}}×\stackrel{→}{\mathrm{B}}=\left[\left(30{\mathrm{v}}_{\mathrm{y}}+40\right)\hat{\mathrm{i}}-\left(30{\mathrm{v}}_{\mathrm{x}}-20\right)\hat{\mathrm{j}}+\left(-20{\mathrm{v}}_{\mathrm{x}}-10{\mathrm{v}}_{\mathrm{y}}\right)\hat{\mathrm{k}}\right]$

Hence,

role="math" localid="1663231379655" $$\begin{gathered} (4.0×{10}^{-17})\hat{\mathrm{i}}+(2.0×{10}^{-17})\hat{\mathrm{j}}=1.6×{10}^{-19}\left[\left(30{\mathrm{v}}_{\mathrm{y}}+40\right)\hat{\mathrm{i}}-\left(30{\mathrm{v}}_{\mathrm{x}}-20\right)\hat{\mathrm{j}}+\left(-20{\mathrm{v}}_{\mathrm{x}}-10{\mathrm{v}}_{\mathrm{y}}\right)\hat{\mathrm{k}}\right] \\ \frac{(4.0×{10}^{-17})\hat{\mathrm{i}}+(2.0×{10}^{-17})\hat{\mathrm{j}}}{1.6×{10}^{-19}}=\left[\left(30{\mathrm{v}}_{\mathrm{y}}+40\right)\hat{\mathrm{i}}-\left(30{\mathrm{v}}_{\mathrm{x}}-20\right)\hat{\mathrm{j}}+\left(-20{\mathrm{v}}_{\mathrm{x}}-10{\mathrm{v}}_{\mathrm{y}}\right)\hat{\mathrm{k}}\right] \\ (2.5×{10}^2)\hat{\mathrm{i}}+(1.25×{10}^2)\hat{\mathrm{j}}=\left[\left(30{\mathrm{v}}_{\mathrm{y}}+40\right)\hat{\mathrm{i}}-\left(30{\mathrm{v}}_{\mathrm{x}}-20\right)\hat{\mathrm{j}}+\left(-20{\mathrm{v}}_{\mathrm{x}}-10{\mathrm{v}}_{\mathrm{y}}\right)\hat{\mathrm{k}}\right] \end{gathered}$$

Thus, $-30v_x+20=1.25×{10}^2$

From part a) ,

$$\begin{gathered} -20v_x-10v_y=0 \\ {v}_y=-2v_x \end{gathered}$$

Hence,

$\begin{array}{rcl}{V}_y& =& -2\left(-3.5\mathrm{x}{10}^3\right)\mathrm{m}/\mathrm{s}\\ & =& 7.0×{10}^3\mathrm{m}/\mathrm{s}\end{array}$

Hence, the value of $v_y$is $7.0×{10}^3\mathrm{m}/\mathrm{s}$.

Therefore, magnetic force depends on charge, velocity of the particle, and applied magnetic field.