A proton traveling at 3.80 km/s suddenly enters a uniform magnetic field of 0.78

ID: 1536964 • Letter: A

Question

A proton traveling at 3.80 km/s suddenly enters a uniform magnetic field of 0.780 T , traveling at an angle of 55.0 o with the field lines

Find the direction of the force this magnetic field exerts on the proton. (Into the page or out)

Part B

Find the magnitude of the force this magnetic field exerts on the proton.

Part C

If you can vary the direction of the proton's velocity, find the magnitude of the maximum and minimum forces you could achieve.

Answer the question in the order indicated. Separate your answers using comma.

Part D

How the velocity should be oriented to achieve the forces in part (C)?

Part E

What would the answers to part (A) be if the proton were replaced by an electron traveling in the same way as the proton?

Please Choose: The force is directed into the page. The force is directed out of the page.

Part F

What would the answers to part (B) be if the proton were replaced by an electron traveling in the same way as the proton?

F = 3.88×1016 N

(B) F = q (v X B) = q v B sin(thheta)

= (1.602 x 10^-19) (3.80 x 10^3) (0.780) sin55

= 3.885 x 10^-16 N

then Fmax = q V B = 4.74 x 10^-16 N

min when theta = 0

then fmin = 0

(D) Fmax when v is perpndicular to B and minimum when V is eaither parallel or antiparallel to B.

Part is not provided.

{ magnitude will be same but direction will get reversed}

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