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

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is