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Question

A positive charged particle of mass m and charge q is released from rest in a region having E=E0^i and B=B0^j from origin. The net force on the particle at any time t if its velocity at this moment is v=vx^i+vz^k :

[E0 and B0 are constants]

A
q(E0B0vz)^i+qB0vx^k
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B
q(E0+B0vz)^i+qB0vx^k
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C
q(E0B0vz)^iqB0vx^k
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D
q(E0+B0vz)^iqB0vx^k
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Solution

The correct option is A q(E0B0vz)^i+qB0vx^k
In this region, the electric field will provide acceleration for the particle in +ve xdirection and once particle has some velocity in xdirection, the magnetic force will act on it in +ve zdirection (^k) as per q(v×B).

The particle will translate in xdirection, and it will rotate in xz plane.

At any time t, its velocity can be written as,

v=vx^i+vz^k

Net force on the particle at this moment will be,

F=Fe+Fm

F=q[E0^i+(vx^i+vz^k)×B0^j]

F=q[E0^i+(B0vx)^k(B0vz)^i]

F=q(E0B0vz)^i+qB0vx^k

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