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प्रश्न
Protons with kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field, so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field.
उत्तर
Given:
Kinetic energy of proton = K
Distance of the target from the accelerator = l
Therefore, radius of the circular orbit ≤ l
As per the question, the beam is bent by a perpendicular magnetic field.
We know
`r = (mv)/(eB)`
For a proton, the above equation can be written as:
`l = (m_pv)/(eB)` (As r=l)....(i)
Here,
mp is the mass of a proton
v is the velocity
e is the charge
B is the magnetic field
`1/2 m_pv^2 = K`
⇒ `v= sqrt((2K)/m_p`
putting the value of V in the equation (i),we get
`l =( sqrt2K_mp)/(eB)`
`B = sqrt(2Kmp)`
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