JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 18)
A particle of mass m and charge q has an initial velocity $$\overrightarrow v = {v_0}\widehat j$$
. If an electric field $$\overrightarrow E = {E_0}\widehat i$$
and
magnetic field $$\overrightarrow B = {B_0}\widehat i$$
act on the particle, its speed will double after a time:
$${{3m{v_0}} \over {q{E_0}}}$$
$${{\sqrt 2 m{v_0}} \over {q{E_0}}}$$
$${{\sqrt 3 m{v_0}} \over {q{E_0}}}$$
$${{2m{v_0}} \over {q{E_0}}}$$
Explanation
Electric field will increase the speed of particle in x direction.
Fx = qE
$$ \therefore $$ a = $${{qE} \over m}$$
Also vx = at = $${{qE} \over m}$$t
$$v_x^2 + v_y^2 = {v^2}$$
$$ \Rightarrow $$ $$v_x^2 + v_0^2 = {\left( {2{v_0}} \right)^2}$$
$$ \Rightarrow $$ vx = $$\sqrt 3 $$v0
$$ \therefore $$ $${{qE} \over m}$$t = $$\sqrt 3 $$v0
$$ \Rightarrow $$ t = $${{\sqrt 3 m{v_0}} \over {q{E_0}}}$$
Fx = qE
$$ \therefore $$ a = $${{qE} \over m}$$
Also vx = at = $${{qE} \over m}$$t
$$v_x^2 + v_y^2 = {v^2}$$
$$ \Rightarrow $$ $$v_x^2 + v_0^2 = {\left( {2{v_0}} \right)^2}$$
$$ \Rightarrow $$ vx = $$\sqrt 3 $$v0
$$ \therefore $$ $${{qE} \over m}$$t = $$\sqrt 3 $$v0
$$ \Rightarrow $$ t = $${{\sqrt 3 m{v_0}} \over {q{E_0}}}$$
Comments (0)
