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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 parabolic path?


  1. P 2 (ii)

  2. P 2 (iii)

  3. Q 4 (i)

  4. R 2 (iv)


Solution

The correct option is B

P 2 (iii)


For the particle to travel along a parabolic path the angle between velocity and Electric field should not be zero but magnetic field should be zero. 

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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