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Question

A charged particle is projected with velocity v=u^i from the origin, in a region having electric field E and magnetic field B both. Value of B and E are given in column 1 and column 2 respectively and path of the particle is given in column 3.

Column 1 Column 2 Column 3 (Magnetic field) (Electric field) (Path of particle)(i)B=0(i)E=0(P)Circular path(ii)B=B0^i+B0^j(ii)E=E0^i(Q)Helical path(iii)B=B0^j+B0^k(iii)E=B0u(^j^k)(R)Cycloid(iv)B=B0^k(iv)E=E0^k(S)Straight line

In which of the following combination particle moves with uniform speed?


A

(I) (i) (P)

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B

(II) (ii) (S)

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C

(III) (iii) (S)

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D

None of these

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Solution

The correct option is C

(III) (iii) (S)


Force due to magnetic field, FB=q(v×B)

Force due to electric field, FE=qE

Since both these forces are acting on the charged particle and their vector sum is coming out be zero that means no net force is acting on the particle. It will move in a straight line with constant velocity.


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