91Ó°ÊÓ

The total force acting on a proton is given by

$\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathbf{q}\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{+}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{)}\mathbf{=}\mathbf{q}\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{)}$

(a)

The total force acting on a proton is

$\begin{array}{r}\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathbf{q}\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{+}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{)}\mathbf{=}\mathbf{q}\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\mathbf{q}\left(\left[v_x\widehat{i}+v_z\widehat{k}\right]×\left[B_x\widehat{i}\right]\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}{\mathbf{qv}}_{\mathbf{z}}{\mathbf{B}}_{\mathbf{x}}\widehat{\mathbf{j}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}}\\ \mathbf{=}\left(1.60×{10}^{−19}C\right)\left(2.00×{10}^{5}m/s\right)\mathbf{(}\mathbf{0}\mathbf{.500}\mathbf{T}\mathbf{)}\widehat{\mathbf{j}}\mathbf{}\mathbf{}\\ \mathbf{=}\left(1.60×{10}^{−14}N\right)\widehat{\mathbf{j}}\end{array}$

Therefore, the force acting on the proton is$\mathbf{(}\mathbf{1}\mathbf{.60}\mathbf{×}{\mathbf{10}}^{\mathbf{−}\mathbf{14}}\mathbf{N}\mathbf{)}\widehat{\mathbf{j}}$

(b)

The electric field is$\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{=}\left(+2.00×{10}^{4}V/m\right)\widehat{\mathbf{i}}$

Therefore, the force due to the electric field will be at same direction,

So, yes, the proton has a component of acceleration in the direction of the electric field

(c)

The motion is circular due to the magnetic field and the magnetic field is perpendicular to the electric field,

Therefore, the motion will be circular but in x direction.

So, the path of the proton is helix.

(d)

The period of motion is

$\begin{array}{r}\mathbf{T}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€}}{\mathbf{Ó\neg }}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{Ï€³¾}}{\mathbf{\left|}\mathbf{q}\mathbf{\right|}\mathbf{B}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{Therefore}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{T}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€}\left(1.67×{10}^{−27}\mathrm{kg}\right)}{\left(1.60×{10}^{−19}C\right)\mathbf{(}\mathbf{0}\mathbf{.500}\mathbf{T}\mathbf{)}}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\mathbf{1}\mathbf{.312}\mathbf{×}{\mathbf{10}}^{\mathbf{−}\mathbf{7}}\mathbf{s}\\ \mathbf{And}\mathbf{}\mathbf{we}\mathbf{}\mathbf{know}\mathbf{}\mathbf{that}\mathbf{}\mathbf{acceleration}\mathbf{}\mathbf{is}\\ {\mathbf{a}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{eE}}{\mathbf{m}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

And the formula of the displacement will be

$\mathbf{x}\mathbf{−}{\mathbf{x}}_{\mathbf{0}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}\mathbf{x}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{a}}_{\mathbf{x}}{\mathbf{t}}^{\mathbf{2}}$

Put the values in the formula

$\begin{array}{r}\mathbf{x}\mathbf{−}{\mathbf{x}}_{\mathbf{0}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}\mathbf{x}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{a}}_{\mathbf{x}}{\mathbf{t}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{v}}_{\mathbf{0}\mathbf{x}}\mathbf{T}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\left(\frac{F_E}{m}\right){\mathbf{T}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(1.50×{10}^{5}m/s\right)\left(1.312×{10}^{−7}s\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\left(\frac{\left(1.60×{10}^{−19}C\right)\left(2.00×{10}^{4}V/m\right)}{1.67×{10}^{−27}\mathrm{kg}}\right){\left(1.312×{10}^{−7}s\right)}^{\mathbf{2}}\\ \mathbf{=}\mathbf{1}\mathbf{.40}\mathbf{×}{\mathbf{10}}^{\mathbf{−}\mathbf{2}}\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{.40}\mathbf{cm}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, the x-component of the displacement of the proton from its position at t = 0 is 1.40cm