1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Maslowich
3 years ago
12

A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not cons

tant across the cross section of the wire, but rather varies as J=αrJ=αr, where αα is a constant.
(a) By the requirement that J integrated over the cross section of the wire gives the total current I, calculate the constant αα in terms of I and R.
(b) Use Ampere’s law to calculate the magnetic field B(r) for (i) r≤Rr≤R and (ii) r≥Rr≥R. Express your answers in terms of I.
Physics
1 answer:
igor_vitrenko [27]3 years ago
3 0

Answer: (a) α = \frac{3I}{2.\pi.R^{3}}

(b) For r≤R: B(r) = μ_0.(\frac{I.r^{2}}{2.\pi.R^{3}})

For r≥R: B(r) = μ_0.(\frac{I}{2.\pi.r})

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

I_{c} = \int {J} \, dA

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr

I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr

I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}

\alpha = \frac{3I}{2.\pi.R^{3}}

For these circunstances, α = \frac{3I}{2.\pi.R^{3}}

(b) <u>Ampere's</u> <u>Law</u> to calculate magnetic field B is given by:

\int\ {B} \, dl = μ_0.I_{c}

(i) First, first find I_{c} for r ≤ R:

I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr

I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr

I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr

I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}

I_{c} = \frac{I.r^{3}}{R^{3}}

Calculating B(r), using Ampere's Law:

\int\ {B} \, dl = μ_0.I_{c}

B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} ).μ_0

B(r) = (\frac{Ir^{3}}{R^{3}2.\pi.r}).μ_0

B(r) = (\frac{Ir^{2}}{2.\pi.R^{3}} ).μ_0

For r ≤ R, magnetic field is B(r) = (\frac{Ir^{2}}{2.\pi.R^{3}} ).μ_0

(ii) For r ≥ R:

I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr

So, as calculated before:

I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}

I_{c} = I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = (\frac{I}{2.\pi.r} ).μ_0

For r ≥ R, magnetic field is; B(r) = (\frac{I}{2.\pi.r} ).μ_0.

You might be interested in
A player bounces a basketball on the floor, compressing it to 80.0 % of its original volume. The air (assume it is essentially N
son4ous [18]

Answer: 292.95 J

Explanation:

change in internal energy= Heat transfer - work done

ΔU =Q -PΔV

Here, Q = 0 as there is no heat transfer.

P =2.00 atm = 2.00 × 101235 Pa = 202470 Pa

ΔV = final volume - initial volume = 0.8 V -V = -0.2 V

where V is the initial volume.

Volume of a spherical ball, V = \frac{4}{3}\pi r^3

r = d/2 = 23.9 cm / 2 = 0.12 m

V = \frac{4}{3}\times 3.14 \times (0.12m)^3= 7.23\times10^{-3}m^3

\DeltaU = -P\DeltaV = - 202470 Pa \times -0.2 \times 7.23\times10^{-3}m^3=292.95 J

Hence, internal energy would change by 292.95 J.

3 0
3 years ago
Read 2 more answers
A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the
nasty-shy [4]

Answer:

The average power the student expended to overcome gravity is 560W, Watts is a units of work it is Joules /time or kg*m^{2}/ s^{3}

Explanation:

Weight = 700N

F= m*g*h \\m*g= 700N\\F= 700N*h\\F=700N*8m\\F= 5600 N*m \\F=5600 J

The power is the work in (Joules) or (N*m) in a determinate time (s) to get Watts (W) units for work

Work= \frac{5600 J}{10s} = \frac{5600 \frac{kg*m^{2} }{s^{2} } }{10 s}  \\Work= 560\frac{kg*m^{2} }{s^{3} } \\Work =560 W

3 0
3 years ago
A company randomly selects 100 light bulbs every day for 40 days from its production process. If 600 defective light bulbs are f
lutik1710 [3]

Answer:

3 sigma lower control limit = 0.0429

Explanation:

Given.

n = 100

days = 100

Number of defective bulbs = 600 defective bulbs

Let p = Process Average

p = 600/(100*40)

P = 600/4000

P = 0.15

q = 1 - p

q = 1 - 0.15

q = 0.85

3 sigma lower limit = p - 3*√(pq/n)

Using the above formula

Substitute in the values

3 sigma lower control limit = 0.15 - 3 * √(0.15 * 0.85/100)

3 sigma lower control limit= 0.15 - 3√0.001275

3 sigma lower control limit = 0.15 - 3* 0.035707142142714

3 sigma lower control limit = 0.15 - 0.107121426428142

3 sigma lower control limit = 0.04287857357185

3 sigma lower control limit = 0.0429 ---- approximated

3 0
3 years ago
Read 2 more answers
Which best explains satellites? The moon is a satellite of the Earth, and the Earth is a satellite of the Sun. Mercury and Pluto
expeople1 [14]
A satellite is a body that revolves around another larger body. Where the larger body has a stronger gravitational pull on the smaller body, keeping it in orbit. Thus the moon is the satellite of the Earth is a good example of this.
4 0
3 years ago
There is a force between two charges; if the distance between the two charges is doubled, by what factor does the force between
Luba_88 [7]
It will double it and yes it will change them.
4 0
4 years ago
Read 2 more answers
Other questions:
  • You have a neutral balloon. What is its charge after 12000 electrons have been removed from it? The elemental charge is 1.6 × 10
    11·1 answer
  • Shawn and his bike have a total mass of
    10·1 answer
  • Why do we believe that most of the mass of the milky way is in the form of dark matter?
    6·1 answer
  • Identify each of these substances as acidic, basic, or neutral.
    12·2 answers
  • An object's (mass or weight) will remain constant throughout the universe, but its (mass or weight) can change from planet to pl
    10·2 answers
  • Which of the following statements are true?
    9·2 answers
  • Where are you MOST likely to find platinum? Check all that apply.
    10·1 answer
  • Students hypothesized that by running an electric current through the wire of the apparatus shown here, they could cause a non-m
    7·2 answers
  • What is the current (Amps) of a circuit with a 9V battery and a 18Ω lamp?
    8·1 answer
  • <img src="https://tex.z-dn.net/?f=%5Chuge%5Cmathfrak%7BQuestion%3A-%7D" id="TexFormula1" title="\huge\mathfrak{Question:-}" alt=
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!