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Pavel [41]
3 years ago
9

A uniform non-conducting ring of radius 2.68 cm and total charge 6.08 µC rotates with a constant angular speed of 4.21 rad/s aro

und an axis perpendicular to the plane of the ring that passes through its center. What is the magnitude of the magnetic moment of the rotating ring?
Physics
1 answer:
Harrizon [31]3 years ago
7 0

Answer: 1.72*10^-7

Explanation:

Given

Radius of the ring, r = 2.68 cm = 0.0268 m

Charge on the ring, q = 6.08 µC

Angular speed of the ring, w = 4.21 rad/s

First, we know that the charge per unit area, σ = q / πr²

Also, the charge on ring of width, dr = σ⋅2πrdr

The Magnetic moment of this ring of width dr.dμ = i⋅A

If we integrate dr at R(top) and at 0(bottom), we get

∫dµ = ∫(R, 0) T⋅2πrdr.(w/2π).πr²

On finding at (R, 0), we get

μ = qwR² / 4

On substituting our values, we have

μ = (6.08*10^-6 * 4.21 * 0.0268) / 4

μ = (6.08*10^-6 * 0.113) / 4

μ = 6.87*10^-7 / 4

μ = 1.72*10^-7

The magnitude of the magnetic moment is 1.72*10^-7

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katrin [286]

To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

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Here,

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Replacing we have that,

F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})

F = 3.6956*10^{-9}N

The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

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a_e = \frac{3.6956*10^{-9}}{9.11*10^{-31}}

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The acceleration of the proton is given as,

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3 0
3 years ago
A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp
Lorico [155]

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

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The friction force is

F_f = T_f/R = 0.06615 / 0.15 = 0.441 N

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3 years ago
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Explanation:

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b -  molality, moles of solute per kilogram of solvent.
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