Question
Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed.
✓
Answer
The velocity $\vec{v}$, of the charged particles, and the $\overrightarrow{E}$ and $\overrightarrow{B}$ vectors, should be mutually perpendicular. Also the forces on q, due to $\overrightarrow{E}$ and $\overrightarrow{B}$ must be oppositely directed.
$qE=qvB$
$or\text{ }v=\frac{E}{B}$
Alternate Answer
Force due to electric field $= q\overrightarrow{E}$
Force due to magnetic field $=q\text{ }(\overrightarrow{v}\times\overrightarrow{B} )$
The required condition is
$q\overrightarrow{E}=-q\text{ }(\overrightarrow{v}\times\overrightarrow{B})$
$[or\text{ }\overrightarrow{E}=-(\overrightarrow{v}\times\overrightarrow{B})=(\overrightarrow{B}\times\overrightarrow{v})]$
$qE=qvB$
$or\text{ }v=\frac{E}{B}$
Alternate Answer
Force due to electric field $= q\overrightarrow{E}$
Force due to magnetic field $=q\text{ }(\overrightarrow{v}\times\overrightarrow{B} )$
The required condition is
$q\overrightarrow{E}=-q\text{ }(\overrightarrow{v}\times\overrightarrow{B})$
$[or\text{ }\overrightarrow{E}=-(\overrightarrow{v}\times\overrightarrow{B})=(\overrightarrow{B}\times\overrightarrow{v})]$
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