A beam of electrons whose kinetic energy is K = 4.00×10 4 eV emerges from a thin

Question

A beam of electrons whose kinetic energy is K = 4.00×104eV emerges from a thin-foil "window" at the end of an accelerator tube. There is a metal plate a distance d = 0.12 m from this window and perpendicular to the direction of the emerging beam (see the figure). Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field B such that

Bmin=(2*me*K)/(e^2 * d^2)^0.5; n which m and e are the electron mass and charge. How should B. be oriented? Evaluate for the minimum value of B.

lectron oil Deam windOW Tube Plate A beam of electrons whose kinetic energy is K 4.00x104 eV emerges from a thin-foil "window" at the end of an accelerator tube. There is a metal plate a distance d = 0.12 m from this window and perpendicular to the direction of the emerging beam (see the figure). Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field B such that 8min-(2."me K/ ( e2 d2) 0.5 in which m and e are the electron mass and charge. How should B. be oriented? Evaluate for the minimum value of B.

Explanation / Answer

K = kinetic energy of the beam

me = mass of each electron

v = speed of each electron

kinetic energy of the beam is given as

K = (0.5) me v2

v = sqrt(2K/me)                          eq-1

d = radius

e = charge on the electron

for the beam to avoid the plate

magnetic force = centripetal force

evB = me v2/r

B = me v/(er)

B = me sqrt(2K/me) /(er)

B = sqrt(2Kme)/(ed)

B = sqrt(2Kme/(ed)2)

direction of magnetic field must be outward

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