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Phantasy [73]
1 year ago
11

A laboratory electromagnet produces a magnetic field of magnitude 1.50T . A proton moves through this field with a speed of 6.00

×10⁶ m/s . (a) Find the magnitude of the maximum magnetic force that could be exerted on the proton. Explain.
Physics
1 answer:
hram777 [196]1 year ago
3 0

The magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

The magnitude of the force on a charged particle moving in a magnetic field is given by the formula,

F= qvB

Here q is the charge on proton = 1.6 x 10^-19 C.

v is the velocity with which the particle is moving = 6.00 x 10^6 m/s

And B is the value of the magnetic field = 1.5 T

Putting the given values in the above equation,

F = 1.6 x 10^-19 x 6 x 10^6 x 1.5 = 1.44 x 10^-12 N.

Hence, the magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

To know more about "magnetic force", refer to the link given below:

brainly.com/question/13791875?referrer=searchResults

#SPJ4

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A 0.5-kilogram apple falls from a height of 2 meters to 1.50 meters. Ignoring frictional effects, what is the kinetic energy of
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The  final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

U_i +K_i = U_f + K_f

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U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

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K_f is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

mgh_i + K_i = mgh_f + K_f

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