The device shown above (a spectrometer) accelerates an ion from rest through a p

Question

The device shown above (a spectrometer) accelerates an ion from rest through a potential difference E = 800 V between two parallel plates as shown above. The ion has a charge of magnitude 2e. The ion then enters a region containing a uniform magnetic field B = 0.075 T which is constant in magnitude and perpendicular to the plane of the path of the ion. The particle curves in a semicircular path and strikes a detector as shown above. The mass of the ion is 6.64 X 10-27 kg. Recall that e = 1.602 X 10-19 C.

i) What is the sign of the charge on the ion? How do you know?

ii) In what direction must B point in order to produce the trajectory shown? Explain your answer.

iii) Calculate the speed of the ion as it enters the region of the magnetic field.

iv)What is the magnitude of the force on the ion as it enters the region of the magnetic field?

v) Determine the distance between where the ion enters the region of the magnetic field and the point where it strikes the detector.

Explanation / Answer

i) The charge is positive since it is accelerated towards the negative plate.

ii) By right hand or by cross product of F = q(v x B) we can show that the force on the charge is downward.

iii) KE = W

1/2mv^2 = qV

1/2mv^2 = 2e*V

1/2*6.64*10^-27*v^2 = 2*1.6*10^-19*800    => v= 2.8*10^5 m/s

iv) F= q(v x B) = qvB = (2*1.6*10^-19)( 2.8*10^5)(0.075)= 6.72*10^-15 N

v) In the region,

Fc= F

mv^2/r = qvB

(6.64*10^-27*(2.8*10^5)^2)/r = 6.72*10^-15      => r= 0.077 m = 77mm

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