What is the kinetic energy of the particle at the point it enters the magnetic f

Question

What is the kinetic energy of the particle at the point it enters the magnetic field?
What is the voltage on the left hand plate?
What is the magnitude of the magnetic field required to get the molecule to hit the detector at Location A?
3. (12 pts) The figure to the right shows a simple mass spectrom- eter. A molecule with a positive charge of 4.80×10-19 C and a mass of 5.40x10-24 kg starts at rest at the lefthand plate and is accelerated toward the righthand plate. The speed of the molecule as it passes through the hole in the righthand plate is 790 m/s. The molecule enters a region of uniform magnetic field, follows the semicircular path shown, and strikes a detector at Location A, which is a distance d=2.90 cm from the point where the the particle enters the region of magnetic field.

Explanation / Answer

a)Kinetic energy of the particle as it enters the amgnetic field = 0.5mv^2

=0.5*5.4*10^-24 * (790)^2

=1.68*10^-18 J

b)let the voltage be V.so, potential drop = V

so, qV = increase in KE

or 1.68*10^-18 = V*4.8*10^-19

or V=3.5V

c)let the magnetic field be B.so,

balancing the magnetic force and centrifugal force,

qvB = mv^2/R (where R=0.5d)

or B = mv/(qR)

=5.4*10^-24*790/(4.8*10^-19*0.5*0.029)

=0.612 T

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