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Question

Column IColumn IIColumn IIIP) ¯¯¯v=v0^i1) ¯¯¯¯E=E0^ki) ¯¯¯¯B=B0(^i+^j)Q) ¯¯¯v=v0(^i+^j)2) ¯¯¯¯E=E0^iii) ¯¯¯¯B=B0^kR) ¯¯¯v=v0^j3) ¯¯¯¯E=0iii) ¯¯¯¯B=0S) ¯¯¯v=04) ¯¯¯¯E=E0(^i+^j)iv) ¯¯¯¯B=B0^j

Which of the following combinations should be true for the particle to travel along a straight line?


A

P 3 (ii)

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B

R 2 (ii)

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C

P 4 (iii)

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D

Q 4 (i)

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Solution

The correct option is D

Q 4 (i)


For the particle to travel along a straight line velocity and electric field should be parallel and magnetic field should be either zero or they should all be parallel.

Apply Fleming's left hand rule for the direction of magnetic force and the direction of electrostatic force is along the direction of electric field for positive charge and opposite for negative charge.


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