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

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.