Grade 12th passMagnetism

This question has been asked in AIEEE 2007 . Please answer this question with explanation.

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Naina Singh

9 Years agoGrade 12th pass

1 Answer

deepak

9 Years ago

consider an setup like $\overrightarrow{E}$ along the +ve x direction and $\overrightarrow{B}$ −y axis and the velocity $\overrightarrow{v}$ along the positive direction

force due to $\overrightarrow{E}$ on the particle will be in the +ve x direction =Eq

force due to $\overrightarrow{B}$ on the particle will be in the -ve x direction =qvB sin(90)

if there is no alteration in the path of the particle then F_{due to B}= F_{due to E}.....they can cancle as they are in the opposite directions

=> qvB=Eq

=> v = E/B = EB/B²

we also know that the velocity is in the direction of $\overrightarrow{E}$ × $\overrightarrow{B}$ (vectorially by right hand rule)

hence $\overrightarrow{v}$ = $[\overrightarrow{E} * \overrightarrow{B}] / B^{2}$

hence option d is correct

hope this was helpfull XD

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This question has been asked in AIEEE 2007 . Please answer this question with explanation.

Naina Singh

1 Answer

deepak

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