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Blizzard [7]
4 years ago
10

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 direc

ted horizontally away from the observer. describe the electron's path in this field when viewed from above.
Physics
1 answer:
Serggg [28]4 years ago
7 0
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.
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Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
g= \frac{GM}{r^2} 
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, d, and its volume (the volume of a sphere of radius r):
M=dV=d ( \frac{4}{3} \pi r^3)=  \frac{4}{3} \pi d r^3 

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
g' =  \frac{GM'}{r'^2} (1)
so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius: 
r'=2r (2)
and that its density is 2/3 of Earth's density:
d'= \frac{2}{3} d
so the mass M' of the new planet is, with respect to the Earth's mass:
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And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
g'= \frac{G( \frac{16}{3}M) }{(2r)^2}=  \frac{GM}{r^2}  \frac{4}{3} =  \frac{4}{3}g
And since g=9.81 m/s^2, we find
g'=  \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2
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