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An electron is projected vertically upward with a speed of 1.70 ✕ 106 m/s into a uniform magnetic field of 0.320 t that is directed horizontally away from the observer. describe the electron's path in this field when viewed from above.

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skyluke89
The electron's path in the magnetic field is a straight line when viewed from above.

In fact, the electron initially moves upward, while the magnetic field is directed horizontally. The electron experiences a force due to the magnetic field (the Lorentz force), whose direction is given by the right-hand rule:
- index finger --> initial direction of the electron (upward)
- middle finger --> direction of the magnetic field (horizontally, away from the observer)
- opposite direction to the thumb* --> direction of the force (horizontally, but perpendicular to the magnetic field, to the right)

This means that the Lorentz force makes the electron moving perpendicular to the magnetic field in the horizontal plane, and since the direction of the field is not changing, this force does not change its direction, so the electron moves in the same direction of the force in the horizontal plane (to the right), therefore following a straight line.

* the direction should be reversed because the charge is negative.

The path of the electron is 30.2μm

Data;

  • speed = 1.70x10^6 m/s
  • magnetic field = 0.320T
  • Charge of an electron = 1.60x10^-19 C
  • mass of an electron = 9.1x10^-31 kg

The radius or path of the electron

The path of the electron is circular and the radius can be calculated with the help of centripetal force.

[tex]qvB = \frac{mv^2}{r} \\qB = \frac{mv}{r}\\ r = \frac{mv}{qB} \\[/tex]

Let's substitute the value and solve.

[tex]r = \frac{mv}{qB}\\r = \frac{9.10*10^-^3^ * 1.70* 10^6}{(1.60*10^-^1^9 * 0.320}\\r = 3.02*10^-5\\r = 30.2 \mu m[/tex]

The path of the electron is 30.2μm

Learn more on centripetal force and magnetic force here;

https://brainly.com/question/26740654

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