A particle of charge ‘q’ and mass ‘m’ starts moving from the origin under the action of an electric field E→=2E0i^ and B→=B0i^ with a velocity v→=v0j^ . The speed of the particle will becomes 2v0 after a time.

A
$t = \frac{2 m v_{0}}{q E_{o}}$
B
$t = \frac{2 B q}{m v_{0}}$
C
$t = \frac{\sqrt{3} B q}{m v_{0}}$
D
$t = \frac{\sqrt{3} m v_{0}}{2 q E_{o}}$

Solution:

$\vec{E}$ is parallel to $\vec{B}$ and $\vec{V}$ is perpendicular to both, therefore path of the particle is a helix with increasing pitch .

Speed of particle at any time ‘t’ is given by
$V = \sqrt{V_{x}^{2} + V_{y}^{2} + V_{z}^{2}}$
here, $V_{y}^{2} + V_{2}^{2} = V_{0}^{2}$ at all times because the yz components will remain same.

Only the x component will increase due to electric field.
After some time, it is given that $V = 2 V_{o}$

$U \sin g, V = \sqrt{V_{x}^{2} + V_{y}^{2} + V_{z}^{2}} \Rightarrow 2 V_{o} = \sqrt{V_{x}^{2} + {V_{o}}^{2}} \Rightarrow V_{x} = \sqrt{3} V_{o}$
$E q u a t i o n o f k i n e m a t i c s, V_{x} = U_{x} + a_{x} t \Rightarrow \sqrt{3} V_{0} = \frac{q 2 E_{o}}{M} t$
$\Rightarrow t = \frac{\sqrt{3} M V_{0}}{2 q E_{o}}$ .

A particle of charge ‘q’ and mass ‘m’ starts moving from the origin under the action of an electric field $\vec{E} = 2 E_{0} \hat{i}$ and $\vec{B} = B_{0} \hat{i}$ with a velocity $\vec{v} = v_{0} \hat{j}$ . The speed of the particle will becomes $2 v_{0}$ after a time.

Solution:

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