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Question

(a)

$\overrightarrow{ B }=\frac{\mu_0}{4 \pi} q \frac{\overrightarrow{ v } 	imes \overrightarrow{ r }}{ r ^3}$ and $\overrightarrow{ E }=\frac{1}{

Vedclass · Accepted Answer

$\frac{6 \hat{ i }-2 \hat{ j }}{ c ^2}$

Vedclass · Answer

(a)

$\overrightarrow{ B }=\frac{\mu_0}{4 \pi} q \frac{\overrightarrow{ v } 	imes \overrightarrow{ r }}{ r ^3}$ and $\overrightarrow{ E }=\frac{1}{4 \pi \epsilon_0} \frac{ q \overrightarrow{ r }}{ r ^3}$

$\mu_0 \in_0(\overrightarrow{ v } 	imes \overrightarrow{ E })=\frac{\overrightarrow{ v } 	imes \overrightarrow{ E }}{ c ^2}=\frac{(\hat{ i }+3 \hat{ j }) 	imes 2 \hat{ k }}{ c ^2}$

$=\frac{-2 \hat{ j }+6 \hat{ i }}{ c ^2}=\frac{6 \hat{ i }-2 \hat{ j }}{ c ^2}$

$\left[c=\frac{1}{\sqrt{\mu_0 \epsilon_0}} \Rightarrow \mu_0 \in_0=\frac{1}{c^2}ight]$

A particle is moving with velocity $\overrightarrow{ v }=\hat{ i }+3 \hat{ j }$ and it produces an electric field at a point given by $\overrightarrow{ E }=2 \hat{ k }$. It will produce magnetic field at that point equal to (all quantities are in SI units)

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