part 1 part 2 part 3 5.0 Points The strengths of the electric and magnetic field

Question

part 1

part 2

part 3

5.0 Points The strengths of the electric and magnetic fields inside the velocity selector are respectively E = 500 V/m and B1= 0.0200 T. What is the speed of the ions that are selected? A. 4.00 × 10-5 m/s B. 5.00 × 102 m/s C. 2.50 × 105 m/s D. 10.0 m/s E. 2.50 × 104 m/s Reset Selection

part 2

5.0 Points The strength of the magnetic field inside the mass spectrometer is B2 = 5.00 × 10?3 T. If the radius of the trajectory of the ion is found to be r = 0.626 m, what was the charge to mass ratio q/m? A. 5.01 × 10?3 C/kg B. 2.00 × 106 C/kg C. 3.13 × 106 C/kg D. 7.83 × 105 C/kg E. 7.99 × 106 C/kg Reset Selection

part 3

5.0 Points From the figure, can you tell whether the ion was positively charged or negatively charged? If yes, how? A. The ion was negatively charged. We can tell this from the direction that the ion curves in the mass spectrometer. B. The ion was negatively charged. We can tell this from the fact that the ion came straight through the velocity selector. C. The sign of the ion's charge cannot be determined from this experiment. D. The ion was positively charged. We can tell this from the fact that the ion came straight through the velocity selector. E. The ion was positively charged. We can tell this from the direction that the ion curves in the mass spectrometer.

Explanation / Answer

F = q( E + V*B) = 0 so

velocity = E/B = 500 / 0.02 = 2.5*10^4 m/s

we know

r= mv/qB

so

q/m = v/rB = 2.5*10^4 / ) 0.626*5*10^-3)

= 7.99*10^6 C/kg

part 3

your diagram is not very clear

send send it to lalitkn@iitk.ac.in will solve then part 3

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