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A particle is moving with velocity i+3j  and it produces an electric field 2k at a point given by . It will produce magnetic field at that point equal to  (all quantities are in S.I. units)

 

Option: 1

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


Option: 2

\frac{6\hat{i}+2\hat{j}}{c^{2}}


Option: 3

 zero


Option: 4

cannot be determined from the given data


Answers (1)

best_answer

 

In Vector Form -

- wherein

 

 

magnetic field B can be written as    \vec{B}=\frac{\mu _{0}}{4\pi }q\frac{\vec{v}\times\vec{r}}{r^{3}}\; and\; \vec{E}=\frac{1}{4\pi \epsilon _{0}}\; \frac{q\vec{r}}{r^{3}}

\therefore \; \vec{B}=\mu _{0}\epsilon _{0}\left ( \vec{v}\times \vec{E} \right )=\frac{\vec{v}\times\vec{E}}{c^{2}}.

(i+3j) \times 2k/c^2=\frac{6i-2j}{c^2}

Posted by

Kuldeep Maurya

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