A uniform magnetic field of 0.25 T points vertically downward. A proton enters the field with a horizontal velocity of 4.0 10 m/s north. What are the magnitude and direction of the instantaneous force exerted on the proton as it enters the magnetic field

Correct question: A uniform magnetic field of 0.25 T points vertically downward. A proton enters the field with a horizontal velocity of 4.0×10⁻² m/s north. What are the magnitude and direction of the instantaneous force exerted on the proton as it enters the magnetic field

1.602×10⁻²¹ N west

Explanation:

The force on a charge moving in a magnetic field is given as,

F = Bqv Equation 1

Where F = force on the proton, B = magnetic Field, q = charge of the proton, v = velocity of the proton.

Given: B = 0.25 T, v = 4.0×10⁻² m/s, q = 1.602 x 10⁻¹⁹ C

Substitute into equation 2

F = 0.25(4.0×10⁻²)(1.602 x 10⁻¹⁹)

F = 1.602×10⁻²¹ N

Hence the magnitude and direction of the instantaneous force = 1.602×10⁻²¹ N west

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.

f=ma newton

f=100x2.5=250n