Physics, 16.10.2020 15:01 ramentome7542

show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through this same potential after the proton crosses into the region with the magnetic field.

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Answer from: jwbri

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

(a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

Explanation: (a) Potential difference is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = -109.44 x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is -109.44 x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$

$qvB=\frac{mv^{2}}{r}$

$r=\frac{mv}{qB}$

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

Answer from: Quest

c. magnitude and direction

Answer from: Quest

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf, ℰ, to a battery's terminal voltage, v, current, i, and internal resistance, r, is ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

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