Question
A particle having charge $10^{- 9}C$ moving in $x - y$ plane in fields of $0.4\widehat{j}N/C$ and $4 \times 10^{- 3}kT$ experiences a force of $(4\widehat{i} + 2\widehat{j}) \times 10^{- 10}\text{ }N$. The velocity of the particle at that instant is $\_\_\_\_$ $m/s$.
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Solution
Sol. [Key Concept: $F = qE + q(v \times \ B)$; equate the components to solve for the velocity.]
$$\overrightarrow{F} = q\overrightarrow{E} + q(\overrightarrow{V} \times \overrightarrow{B})$$
$$\begin{matrix} & \ (4\widehat{i} + 2\widehat{j}) \times 10^{- 10} \\ = & 10^{- 9}\left\lbrack 0.4\widehat{j} + \left( v_{x}\widehat{i} + v_{y}\widehat{j} \right) \times \left( 4 \times 10^{- 3} \right)\widehat{k} \right\rbrack \\ & \ (4\widehat{i} + 2\widehat{j}) \times 10^{- 10} \\ = & 10^{- 9}\left( 0.4\widehat{j} - 4 \times 10^{- 3}v_{x}\widehat{j} + 4 \times 10^{- 3}v_{y}\widehat{i} \right) \end{matrix}$$
On comparing
$${v_{y} = 100,v_{x} = 50 }{\overrightarrow{V} = 50\widehat{i} + 100\widehat{j} }$$
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