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# A proton travels with a speed of 4.5×106 m/s at an angle of 120◦ west of north. A magnetic PHYSICS HELP PLEASE?

1) A proton travels with a speed of 4.5×106 m/s

at an angle of 120◦ west of north. A magnetic

field of 4.3 T points to the north.

Find the magnitude of the magnetic force

on the proton. (The magnetic force experienced by the proton in the magnetic field is

proportional to the component of the proton’s

velocity that is perpendicular to the magnetic

field.)

Answer in units of N.

2)What is its direction?

1. East

2. North

3. out of the Earth

4. West

5. into the Earth

6. South

7. None of these

3)Find the magnitude of the proton’s acceleration as it moves through the magnetic field.

Answer in units of m/s^2

please explain each step because i dont understand anything. not even formulas

### 1 Answer

- NCSLv 710 months agoFavorite Answer
1/ vector F = q * v x B

where v and B are vectors and "x" represents the cross product.

I like to convert to "standard" -- measured ccw from East; so Θ = 210º

then

F = 1.6e-19C * 4.5e6m/s * (cos210º i + sin210 j) x 4.3T j

where i and j are the unit vectors.

so

F = -2.68e-12 N

and so the magnitude is 2.68e-12 N ◄

2/ the sign indicates "down," or into the Earth ◄ 5

3/ a = F / m = 2.68e-12N / 1.67e-27kg = 1.61e15 m/s²

comes from Newton's Second: F = ma

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