A charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields ${\bf{E}}$ and ${\bf{B}}$ with a velocity ${\bf{v}}$ perpendicular to both ${\bf{E}}\;and\;{\bf{B}}$, and comes out without any change in magnitude or direction of ${\bf{v}}$. Then

${\bf{v}}\; = \;{\bf{E}}\; \times \;{\bf{B}}/{{\rm{B}}^2}$

${\bf{v}}\; = \;{\bf{B}}\; \times \;{\bf{E}}/{{\rm{B}}^2}$

${\bf{v}}\; = \;{\bf{E}}\; \times \;{\bf{B}}/{{\rm{E}}^2}$

${\bf{v}}\; = \;{\bf{B}} \times \;{\bf{E}}/{{\rm{E}}^2}$

NEET UG PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

1 Medium

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Application +4 | -1

A charged particle with charge

q

enters a region of constant, uniform and mutually orthogonal fields

\mathbf{E}

and

\mathbf{B}

with a velocity

\mathbf{v}

perpendicular to both

\mathbf{E}\;and\;\mathbf{B}

, and comes out without any change in magnitude or direction of

\mathbf{v}

. Then

View answer and explanation

Correct option: A

\mathbf{v}\; = \;\mathbf{E}\; \times \;\mathbf{B}/{\mathrm{B}^2}

Option A is correct because the condition for a charged particle to move undeflected in mutually perpendicular electric and magnetic fields is

\mathbf{v} = \mathbf{E} \times \mathbf{B}/{\mathrm{B}^2}

Practice All Chapter MCQs →

5 \mathrm{~cm}

12 \mathrm{~cm}

250 \mu \mathrm{T}

d \vec{B}

d \vec{l}

\vec{r}

i

\oint {\overrightarrow B .\overrightarrow {dl} }