I need help in doing this problem for (a) I tried this method but it was wrong I

Question

I need help in doing this problem for (a)

I tried this method but it was wrong

I need help

7. Question Details Tiplere 26.P037 The plates of a Thomson q/m apparatus are 6.0 cm long and are separated by 1.6 cm. The end of the plates is 30.0 cm from the tube screen. The kinetic energy of the electrons is 2.6 keV. (a) If a potential of 25.0 Vis applied across the deflection plates, by how much will the beam deflect? 9.59 X mma (b) Find the magnitude of the crossed magnetic field that will allow the beam to pass through undeflected. 5.17e-05 T eBook Submit Answer Save Progress Practice Another Version My Notes

Explanation / Answer

a) Since K = 2.6keV = 2.6*1.60x10^-16J/keV = 4.16x10^-16 => v = sqrt(2*K/m)

= sqrt(2*4.16x10^-16/9.11x10^-31) = 3.022x10^7m/s

The time between the plates is t = l/v = 0.060m/3.022x10^7 = 1.985x10^-9s

The acceleration is F/m = E*q/m = V*q/(d*m) = 25*1.60x10^-19/(0.016*9.11x10^-31) = 2.7442x10^14m/s^2

so y = 1/2*a*t^2 = 1/2*2.7442x10^14*(1.985x10^-9)^2 = 5.4063x10^-4m (This is the deflections at the end of the plates)

Now the particle leaves the plates with vx = 3.022x10^7m/s
and vy = a*t = 2.7442x10^14*1.985x10^-9 = 5.447x10^5m/s

so the angle is arctan( 5.447x10^5/3.022x10^7) = 1.0326o

So the drop over the 30cm is y = 0.30*tan(1.0326) = 5.407x10^-3m

So overall the particle moves 5.4063x10^-4m + 5.407x10^-3 = 5.947x10^-3m (This is the deflection at the screen)


b) A velocity selector comes from q*v*B = E*q => v = E/B

so B = E/v = 25.0V/(0.016m*3.022x10^7) = 5.1704x10^-5T

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.