A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of `10^(5)m//s`.The velocit
A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of `10^(5)m//s`.The velocity is perpendicular to both the fields.When the electric field is swithed off.the proton moves along a circle of radius `2 cm`.Find the magnitudes of electric and the magnitic fields.Take the mass of the proton`=1.6xx10^(-27) kg`.
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Correct Answer - A::B::C
`V=E/B`
`r=(mV)/(qB)`
`B=(mV)/(rq)`
`=(1.6xx10^(-27)xx2xx10^(5))/(4xx10^(-2)xx1.6xx10^(-19))=5x10^(-2)T`
`E=VB=5xx10^(3)N//C`
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