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torisob [31]
3 years ago
13

A proton moves in a circular path perpendicular to a constant magnetic field. If the field strength of the magnet is increased,

does the diameter of the circular path increase, decrease, or remain the same?
Physics
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

The diameter of the circular path decrease

Explanation:

The magnitude of the magnetic force exerted on the proton is given by:

 F=qvB

According to Newton's second law, this force is equal to the centripetal force. Therefore, we have:

F=F_c\\qvB=ma_c

Recall that a_c=\frac{v^2}{r}. Replacing and solving for B:

qvB=m\frac{v^2}{r}\\B=\frac{mv}{qr}

Thus, the field stregth is inversely proportional to the radius of the path and since the radius is directly proportional to the diameter, if the field strength increases, the diameter decreases.

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Hi there!

The mechanical energy of the rider is calculated as the sum of the gravitational potential energy plus the kinetic energy. Since there are no dissipative forces (like friction), the mechanical energy of the rider at a height of 55.0 m above the sea level will be the same at a height of 25.0 m (or at any height), because the loss in potential energy will be compensated by a gain in kinetic energy, according to the law of conservation of energy.

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This mechanical energy is constant because when the rider coast down the hill, its potential energy is being converted into kinetic energy, so that the sum of potential energy plus kinetic energy remains constant.

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