A beam of protons is accelerated easterly from rest through a potential differen

Question

A beam of protons is accelerated easterly from rest through a potential difference of 4.0 . It enters a region where there exists an upward pointing uniform electric field. This field is created by two parallel plates separated by 12 with a potential difference of 250 across them.
a)What is the speed of the protons as they enter the electric field?
Express your answer using two significant figures.
b)Find the magnitude of the magnetic field (perpendicular to ) needed so the beam passes undeflected through the plates.
c)What is the direction of this magnetic field?
toward the south
toward the north

d)What happens to the protons if the magnetic field is greater than the value found in part B?

the protons will be deflected upward
the protons will be deflected downward

Explanation / Answer

(A) initial velocity   u = 0.0 m/s Charge of the proton   q = 1.6*10^-19 C mass of the proton =   1.67*10^-27 kg Potential difference V   = 4.0 V According to law of conservation of energy (1/2)mv^2    =   Vq    sloving for velocity as the proton enters the electric field region v = sqrt ( 2Vq / m )   plug all values and get the answer v = 2.76*10^4 m/s   (or) 27.6*10^3 m/s separaion between the plates   =  0. 12 m (b)  in the secon region electric field E = V / r    = 250 v / 0.12 m   =2083.3 N/C magnetic fleld B = E / v =   2083.3 N/C / 2.76*10^4 m/s    = 0.0755T (c) fleming's left hand rule the direction is towards south (d) As the mangnetic field overcomes the value (b) the proton s will deflects down ward diretion in circular path because the magnetic field dominates the electric force magnetic fleld B = E / v =   2083.3 N/C / 2.76*10^4 m/s    = 0.0755T (c) fleming's left hand rule the direction is towards south (d) As the mangnetic field overcomes the value (b) the proton s will deflects down ward diretion in circular path because the magnetic field dominates the electric force

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