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A charged particle enters into a space a...

A charged particle enters into a space and continues to move undeflected. Considering gravitational force also on the particle, in the space

A

a uniform horizontal electric field and a vertical magnetic field may be present

B

a vertical electric field alone may be present

C

uniform electric field and magnetic field both directed vertically downwards may be present

D

a uniform horizontal magnetic field alone may be present

Text Solution

Verified by Experts

The correct Answer is:
B, C, D

If a uniform electric field horizontally, then the particle will experience a force due to that electric field along the horizontal direction and if magnetic field is vertical then the force experienced by the particle due to the magnetic field will also be horizontal. If these two forces are equal and opposite to each other, then they will neutralize each other, but a vertically downward force (weight of the particle) remains unbalanced. Hence the particle will not move undefelected. Hence, the option (a) is wrong. If the electric field is verticle, then the resultant force on the particle will not move underflected. Hence, the option (a) is wrong. If the electric field is vertical, then the resultant force on the particle will be the resultant of `qvec(E) and mvec(g)`. This resultalt can be zero, and if it is zero then particle can move undeflected. Henceoption (b) is correct.
If both electric field and magnetic field are vertical, and if velocity is also vertical, then net force acting on the particle can be only vertical, so the particle can move undefleted (without change in direction) in vertical direction. Hence option (c) is correct.
If a uniform horizontal magnetic field along exists in the space. and the particle is also moving horizontally but perpendicular to the magnetic field then force exerted by the magnetic field will be vertical. If this force is vertically upwards and is equal to weight of the particle then the resultant force on the particle will be zero. hence, the particle will continue to move with constant velocity. Hence option (d) is correct.
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