A proton traveling at 3.70 km/s suddenly enters a uniform magnetic field of 0.77

Question

A proton traveling at 3.70 km/s suddenly enters a uniform magnetic field of 0.770 T , traveling at an angle of 55.0 with the field lines (see (Figure 1) ).

Part A

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

Part B

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

Express your answer in newtons to three significant figures.

Part C

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

Express your answer in newtons to three significant figures.

Part D

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

Express your answer in newtons to three significant figures.

Part E

How should the velocity be oriented to achieve the forces in Parts C and D.

How should the velocity be oriented to achieve the forces in Parts C and D.

Part F

What would the answer to Part A be if the proton were replaced by an electron traveling in the same way as the proton?

Please ChooseThe force is directed into the page. The force is directed out of the page.

What would the answer to Part B be if the proton were replaced by an electron traveling in the same way as the proton?

Express your answer in newtons to three significant figures.

Explanation / Answer

given, speed of proton, v = 3.7 km/s = 3700 m/s

magnetic field B = 0.77 T

angle with magnetic field, theta = 55 deg

A. the direction of the force will be into the page ( from the left hand rule)

B. Magnitude of this force = qvBsin(theta) = 1.6*10^-19*3700*0.77*sin(55) = 3.734*10^-16 N

C. the mximum magnitude of the force for varying theta will be qvB = 1.6*10^-19*3700*0.77 = 4.5584*10^-16 N

D. the minimum for ce achievable for varying theta is for theta = 0 deg, F = 0 N

E. Part C. theta = 90 deg, F is maximum when v is perpendicular to B

Part D. theta = 0 deg. F is minimum when v is parallel to B

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