Question

An alpha particle, which contains two protons and two neutrons (i.e., charge of +2e and mass of 4m_p), is accelerated from rest through a potential difference of Delta V = -1000 V prior to entering a region of space that contains a uniform magnetic field of 1.00 mT into the page. Determine the speed of the particle as it enters the magnetic field region. Calculate the force in vector notation that the alpha particle experiences just after it enters the magnetic field region. How far does the particle penetrate into the field-filled region noting that its path is a semicircle, i.e., what is the radius of its motion? Determine the time interval the alpha particle spends in the field-filled region, noting that its path is a semicircle. Determine the particle's velocity in vector notation just after it leaves the field-filled region. In which direction would you apply an electric field so that the alpha particle passes through the field-filled region shown without being deflected? Calculate the electric field required for the alpha particle to pass through the field-filled region without being deflected.

Answer 1

In moving throgh potential difference

work done done = change in KE

-delta_V*q = (1/2)*m*v^2

1000*2*1.6*10^-19 = (1/2)*4*1.6*10^-27*v^2

v = 3.16*10^5 m/s

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(b)

B = 1*10^-3 -k

v = 3.16*10^5 i m/s

Fb = q*(vxB)

Fb = 2*1.6*10^-19*( 3.16*10^5 i x 1*10^-3 -k )

Fb = 2*1.6*10^-19*3.16*10^5*1*10^-3 -(ixk)

Fb = 1.011*10^-16 N j

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(c)

In the magnetic field

Fb = Fc

Fc = mv^2/r

1.011*10^-16 = 4*1.67*10^-27*(3.16*10^5)^2/r

r = 6.6 m

distance = pi*r = pi*6.6 = 20.73 m

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(d)

t = distance/v = 20.73/(3.16*10^5) = 6.56*10^-5 s

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(e)

v = - 3.16*10^5 i

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f)

Fb is in down wards

so electric force Fe is upwards

for positive charge electric force is in the direction of electric field

Electric field is in upwards

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g)

Fe = E*q

E = electric field

Fb = Fe

1.011*10^-16 = E*2*1.6*10^-19

E = 316 N/C

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