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When a proton is released from rest in a...

When a proton is released from rest in a room, it starts with an initial acceleration `a_(0)` towards west. When it is projected towards north with a speed `v_(0)` it moves with an initial acceleration `3a_(0)` towards west. The electric and magnetic fields in the room are:

A

`(ma_(0))/(e )` east, `(3ma_(0))/(ev_(0))` down

B

`(ma_(0))/(e )` west `(2ma_(0))/(ev_(0))`up

C

`(ma_(0))/(e )` west, `(2ma_(0))/(ev_(0))` down

D

`(ma_(0))/(e )` east `(3ma_(0))/(ev_(0))` up

Text Solution

Verified by Experts

The correct Answer is:
C
In this question, the proton moves from rest towards west. It is due to a force on proton by virtue of electric field along west.
Acceleration of proton due to electric field `=(eE)/(m)=a_(0) or E=(ma_(0))/(e )` west
When proton projected towards north with a speed `v_(0)`, it moves with acceleration `3a_(0)` towards west, shows that the proton is experiencing forces due to electric field along west and magnetic field acting vertically downwards. Therefore, the acceleration of proton due to magnetic field.
`=3a_(0)-a_(0)=2a_(0)`
Force on proton due to magnetic field
`m(2a_(0))=ev_(0)B or B=(2ma_(0))/(ev_(0))` downwards
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