contestada

A proton moving at 7.0 × 10^4 m/s horizontally enters a region where a magnetic field of 0.10 t is present, directed vertically downward. What magnitude force acts on the proton due to this field?

Respuesta :

asuwafohjames

The force on the proton due to the magnetic field is 1.12×10⁻¹⁵ N.

What is force?

Force can be defined as the product of mass and acceleration

To calculate the force that acts on the proton due to the field, we use the formula below.

Formula:

  • F = qVB.............. Equation 1

Where:

  • F = Force on the proton
  • q = Charge of the proton
  • V = Velocity of the proton
  • B = Magnetic field.

From the question,

Given:

  • q = 1.60 x 10⁻¹⁹ C
  • B = 0.1 T
  • V = 7×10⁴ m/s

Substitute these values into equation 1

  • F = (1.60 x 10⁻¹⁹)(0.1))(7×10⁴)
  • F = 1.12×10⁻¹⁵ N

Hence, the force on the proton due to the magnetic field is 1.12×10⁻¹⁵ N.

Learn more about force here: https://brainly.com/question/25239010

#SPJ1

Otras preguntas