Physics, 08.04.2020 00:12 trin520

An electron and a proton, moving side by side at the same speed, enter a 0.020-T magnetic field. The electron moves in a circular path of radius 7.0 mm. Describe the motion of the proton qualitatively and quantitatively

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Answer from: LordYoshee8676

a. Radius of the proton (r) = 12.85m

b. Velocity of the proton = 24.62 × 10⁶ m/s

c. Periodic Time (T) = 3.22 × 10⁻⁶ s

Explanation:

The motion of a proton can be described quantitatively and qualitatively by finding the velocity of the proton, the periodic time and radius of the proton.

The first step is to find the Radius of the circular path.

We would derive this formula for Force

Force = mass × acceleration

Where acceleration = (velocity)²/ radius

Therefore

Force of a proton = Force of an electron

(mass of a proton × (velocity)²)/radius of a proton = (mass of a electron × (velocity)²)/radius of a electron

Radius of a proton = (mass of a proton × radius of an electron) ÷ mass of and electron

Mass of a proton = 1.67 × 10⁻²⁷kg

Mass of an electron = 9.1 × 10⁻³¹kg

Radius of an electron = 7.0mm = 7.0 × 10m

Radius of a proton = ( 1.67 × 10⁻²⁷kg× 7.0 × 10⁻³m) ÷ 9.1 × 10⁻³¹kg

Radius of a proton = ( 1.67 × 10⁻²⁷kg × 7.0 × 10⁻³m) ÷ 9.1 × 10⁻³¹kg

Radius of a proton = 12.85m

b. Velocity of a proton =

v = (B × r × q) ÷ m

Where B = Magnetic Flux = 0.020T

r = radius of a proton = 12.85m

q = charge of a proton = 1.602 × 10⁻¹⁹ coulombs

m = mass of a proton = 1.67 × 10⁻²⁷kg

Velocity of a proton = ( 0.020T × 12.85m × 1.602 × 10⁻¹⁹c ) ÷ 1.67 × 10⁻²⁷kg

Velocity of a proton = 24.62 × 10⁶ m/s

c. Periodic Time(T) = (2πr) ÷ v

where r = radius of a proton = 12.85m

v = velocity of a proton = 24.62 × 10⁶ m/s

T = (2 × π × 12.85m) ÷ 24.62 × 10⁶ m/s

= 3.22 × 10⁻⁶ s

Answer from: Quest

false.

explanation:

this statement is false. after falling for 5 seconds, its speed will not be 100 m/s.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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