What is the magnetic force on a proton that is moving at 2.5 x 10 m/s to the
left through a magnetic field that is 3.4 T and pointing toward you? The
charge on a proton is 1.6 10-19 c. Use F-qvx B sin(o).

C. 1.4 10-11 N up

step-by-step explanation:

Explanation:

The magnetic force, F on a charge q moving with velocity v in a magnetic field B at an angle θ is given by:

F = q v B sin θ

Charge of proton, q = 1.6 × 10⁻¹⁹ C

Strength of magnetic field, B = 3.4 T pointing outwards

velocity of the proton, v = 2.5 × 10⁷ m/s towards left

Magnetic force is given by:

F =  1.6 × 10⁻¹⁹ C× 2.5 × 10⁷ m/s ×3.4 T× sin 90 = 13.6 × 10⁻¹² N = 1.4 × 10⁻¹¹ N up

The direction of the force is given by Lorentz Right hand rule. The fingers point magnetic field, the thumb points towards velocity, then the force on the proton is given by the direction perpendicular to the palm.  

The magnetic field acts outwards with velocity of the proton towards left. The force would act perpendicular to the two -upwards.

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.

-1/2 +2

step-by-step explanation: