30.10.2 A proton travels with a speed of 8.34x109 m/s at an angle of 32.3 with t

Question

30.10.2 A proton travels with a speed of 8.34x109 m/s at an angle of 32.3 with the 9.00direction of a magnetic field. The field exerts a force of 3.00x10-17 N on the proton. What is the strength of the field? (1/5 submissions used) T Save 30.10.2 Submit 30.10.2 Section 12: Physics at work: velocity selector 30.12.1 The electric field in a velocity selector is generated by two parallel plates (5.00) separated by 5.00 cm with a potential difference of 90.0 V. If the magnetic field is 5.000e-2 T, at what speed will particles pass through undeflected? 0/5 submissions used) m/s Save 30.12.1Submit 30.12.1 30.12.2 A velocity selector is tuned to let charges with a speed of 325 m/s pass through 5.00) If the strength of the magnetic field is 0.250 T, what is the strength of the electric iieeki? (0/5 submissions used) N/C Save 30.12.2 Submit 30.12.2

Explanation / Answer

30.10.2 : F = q (v x B)

F = q v B sin(theta)

3 x 10^-17 = (1.6 x 10^-19) ( 8.34 x 10^5) (B) sin32.3

B = 4.63 x 10^-4 T

30.12.1: E = V / d = (90) / 0.05 = 1800 V/m

B = 5 x 10^-2 T

E = v B

v = (1800) / (5 x 10^-2)

v = 36000 m/s

30.12.2: E = v B

E = (325) (0.250) = 81.25 N/C

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v = E / B = (9.1 x 10^3) / 0.50

= 18200 m/s

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q v B = m v^2 / r

r = m v / q B

r = (3.1 x 10^-6) (59) / (23 x 10^-6 x 8.3)

r = 0.96 m

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