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Question
When a proton is released from rest in a room, it starts with an initial acceleration a0towards west. When it is projected towards north with a speed v0, it moves with an initial acceleration 3a0 towards west. Find the electric field and the maximum possible magnetic field in the room.
Solution
Given:
The initial acceleration of a proton, when it is released from rest, is a0 towards west.
F = qE ....(i)
F = ma0 ....(ii)
Here, q is the charge, E is the electric field and m is the mass.
On equating both the forces F of equations (i) and (ii), we get:
qE = ma0
⇒ `E = (ma_0)/(q)` towards west
When the proton is projected towards north with a speed v0, it moves with an initial acceleration 3a0 towards west.
`vecF = q vecv_0 xx vecB`
where B is the magnetic field.
`⇒ B = | vecF|/(qv_0)`
Again, an electric force will act on the proton in the west direction, due to which, an acceleration a0 will act on the proton.Now, as the proton was initially moving with a velocity v, a magnetic force is also acting on the proton.So, the change in acceleration of the proton will be solely due to the magnetic force acting on it.
Change in acceleration towards west due to the magnetic force
= 3a0 − a0 = 2a0
So, F = m2a0
Hence, required magnetic field,
`B = (2ma_0)/(qv_0)`
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