--------- A mass spectrometer is shown below. The magnetic field is denoted by x

Question

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A mass spectrometer is shown below. The magnetic field is denoted by x's, and there is an electric field only in the lower region (indicated by arrows). The electric and magnetic fields are perpendicular to one another. The magnitude of the magnetic field is 0.0350 tesla and the electric field magnitude is 2500 V/m. The particle is a singly charge ion with mass m = 2.18 Times 10-26 kg. Determine the speed of the particle that is required so that it follows a straight-line path in the region with both fields, as shown in the figure. Determine the radius of the curved path in the region with only a magnetic field. What would happen if the particle were travelling slower than the speed found in (a)? What would happen if the particle had twice the mass?

Explanation / Answer

a)

Bqv = Eq

v = E / B = 2500 / 0.0350 = 71428.5 m/s

b)

mv^2 / r = Bqv

r = mv / Bq = 2.18 x 10^(-26) x 71428.5 / (0.0350 x 1)

= 4.44 x 10^(-20) m

c)

r = mv / Bq

So, If particle will travel slower, then the radius of the path will be less and the particle may not enter the electric field as it may hit the horizontal slab.

d)

r = mv / Bq

so, If particle has twice the mass, the radius of the motion will be twice, thus the particle hit the right part of horizontal slab and will not enter the electric field space.

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