Explain why a sound becomes louder as you move closer to the source of the sound.

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

A rope is tied to a tree 4.5 feet from the ground and then run through a pulley hooked to a vehicle 33 feet from the tree. If a

Angelina_Jolie [31]

Answer:

The vehicle displacement is 9.90 feet.

Explanation:

Given that,

Height of tree = 4.5 feet

Distance = 33 feet

According to figure,

We need to calculate the value of l

Using Pythagorean theorem

$l=\sqrt{(33)^2+(4.5)^2}$

We need to calculate the vehicle displacement

Using horizontal component

Vehicle displacement =horizontal component of pulled rope

$Vehicle\ displacement= d\cos\theta$

Where, $\theta$ is angle between rope and ground

d = pulled length of rope

$Vehicle\ displacement=10\times\dfrac{33}{\sqrt{(33)^2+(4.5)^2}}$

$Vehicle\ displacement=9.90\ feet$

Hence, The vehicle displacement is 9.90 feet.

3 0

3 years ago

Why is it important to understand the concept of density ?

N76 [4]

Density applies to many if not all aspects of life. With density you can explain why ice floats. You can explain why oxygen is on the earth, and not floating around in space( or being replaced by another gas). You can also explain why heat rises while cold air sinks.

8 0

3 years ago

A 50.6 g ball of copper has a net charge of 1.6 µc. what fraction of the copper's electrons have been removed? (each copper atom

Fynjy0 [20]

First figure out how many atoms you have with Avogadro's number. Since there are 63.5 grams/mol and you have 50.6 grams, you have (50.6/63.5)6.022E23=4.7986E23 atoms. Since there are 29 protons per atom, there are also 29 electrons per atom, so you should have a total of
29*4.7986E23=1.3916E25 electrons.
Since there is a positive charge you know some of these electrons are missing. How many are missing can be found by dividing the charge you have by the charge on the electron: 1.6E-6/1.6022E-19 = 9.98627E12 electrons are missing.
Now take the ratio of what is missing to what there should be:
9.98627E12/1.3916E25 = 7.1760873E-13

5 0

3 years ago

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .