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3 years ago

How did rutherfords discovery effect the model of the atom

1 answer:

7 0

<h2>Thomson proposed the plum pudding model of the atom, which had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus.</h2>

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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}
$

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

What would happen if there were no friction between the girl and the slope?

erica [24]

She'll most likely fall

7 0

3 years ago

A stone tumbles into a mine shaft and strikes bottom after falling for 4.2 second how deep is the mine shaft

rjkz [21]

$4.2*9.8\\41.16$

41.16 meters

5 0

3 years ago

Find a first-degree polynomial function p1 whose value and slope agree with the value and slope of f at x = c. f(x) = cot(x), c

Neporo4naja [7]

The correct answer is y=-2x+(1/2)

y = f'(x)· x + c

Y = -2x + C

1 = -2x π/4 + C

=) C = I + π/2

y=-2x+(1/2) is the first-degree polynomial.

First-degree polynomials are the simplest polynomials. Here, we'll talk about a few qualities and connect the terms polynomial, function, and equation. Write a polynomial equation in standard form before attempting to solve it. Factor it, then set each variable factor to zero after it has reached zero. The original equations' answers are the solutions to the derived equations. Factoring cannot always be used to solve polynomial equations. For instance, the polynomial 2x+5 has an exponent of 1. The most typical kinds of polynomials used in algebra and precalculus are zero polynomial functions.

Learn more about polynomial functions here :-

brainly.com/question/22592200

#SPJ4

7 0

1 year ago

When you have a pot of water on the stove, heat is transferred to the water. Describe the behavior of the water molecules and ho

Digiron [165]

The hotter molecules become, the faster they move around. The colder they are, the more slow and lethargic they are

3 0

2 years ago