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A proton traveling at 3.90km/s suddenly enters a uniform magnetic field of 0.790

ID: 1315998 • Letter: A

Question

A proton traveling at 3.90km/s suddenly enters a uniform magnetic field of 0.790T , traveling at an angle of 55.0o with the field lines

Find the direction of the force this magnetic field exerts on the proton: into or out of the page.

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

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.

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

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

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

Explanation / Answer

Force = q V x B = qVBsin theta

direction will be given by the right hand thumb rule

the direction is into page

put your hand along the velocity and cult your fingers in the direction of magnetic field then direction of the thumb gives direction of the force

F = 1.6 x 10^-19 x 3.9 x [5 / 18 ] x 0.79 sin 55

F = 1.12 x 1-^-19 N

now the force depends on the value of sin theta

if sin theta = 0 i.e theta = 0 then force = 0

so minimum force is when the protom moves parallel to the magnetic field

max force will act when sin theta = 1 i.e theta = 90

max force will act when proton moves perpendicular to the field

if the proton is replaced by an electron only the direction of forces will change to exactly opposite direction

direction of force for proton is out of page

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