Aproton moves around a circular path ( radius = 2.0 mm ) in a uniform .25-t magnetic field. what total distance does this proton travel during a 1.0-s time interval? ( m= 1.67 x 10⁻²⁷ kg = 1.6 x 10⁻¹⁹ c)a) 82 kmb) 59 kmc) 71 kmd) 48 kme) 7.5 km

Option D. 48 km

Explanation:

In order to do this, let's remember something about the magnetic fields:

"Magnetic forces are directed perpendicular to the velocity vector of a charged particle and the magnetic field direction. Since Newton's second law tells us that the direction of motion is the same as the direction of a force, this means that we have perpendicular motion as well. This manifests in the form of circular motion inside the magnetic field. This is called "cyclotron motion".

According to this, let's write the expression of magnetic field due to the cyclotron motion:

mV²/r = qVB (1)

Where:

m: mass of the proton

q: charge of proton

V: speed of the proton

B: magnetic field

r: radius of the path

To calculate the total distance, we use the following expression:

d = V*t (2)

Therefore, in order to calculate distance, we should calculate first the speed of the proton, and then, replace here to solve for the distance.

Now, solving for V in expression (1) we have:

mV²/r = qVB   V cancels out

mV/r = qB

V = qBr/m  (3)

Now, we will replace the given values to solve for V. Remember that radius should be in meter (divide by 100):

V = 1.6x10^-19 * 0.25 * 0.002 / 1.67x10^-27

V = 47,904.19 m/s

Now, let's put this value in equation (2) and solve for distance:

d = 47,904.19 * 1

d = 47,904.19 m

But the answer is expressed in km, so:

d = 47,904.19 / 1000

d = 47.9 km

This is the total distance the proton travels.

b.transmitters send radio waves, and receivers capture radio waves.

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge