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A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (the figure (Figure 1)). The beam travels a distance of 1.60 cm while in the field.

What is the magnitude of the magnetic field?

Respuesta :

onyebuchinnaji

The magnitude of the magnetic field of the given protons is 4.69 x 10⁻⁹ T.

The given parameters;

  • speed of the proton, v = 1.2 km/s = 1200 m/s
  • distance traveled by the beam, s = 1.6 cm = 0.016 m

The magnitude electric force of the given protons is calculated as;

[tex]F = \frac{Kq^2}{r^2}[/tex]

The magnitude of the magnetic force of the given protons is calculated as;

[tex]F = qvB[/tex]

The magnitude of the magnetic field of the given protons is calculated as;

[tex]qvB = \frac{kq^2}{r^2} \\\\B = \frac{kq}{r^2v} \\\\B = \frac{9\times 10^{9} \times 1.6\times 10^{-19}}{(0.016)^2 \times 1200 } \\\\B = 4.69 \times 10^{-9} \ T[/tex]

Thus, the magnitude of the magnetic field of the given protons is 4.69 x 10⁻⁹ T.

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