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3 years ago

Consider first the generation of the magnetic field by the current I1(t)I1(t)I_{1}(t) in solenoid

Physics

1 answer:

densk [106]3 years ago

7 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

So the magnetic field on the solenoid 1 is would be

$B_1 (t) = \mu_0 n_1 I_1 (t)$

Explanation:

Now in this question we are given an solenoid and our interest is on the magnetic field on the solenoid

Now first we will consider the Ampere's law which is mathematically represented as

∮ $\= B \cdot d \=s = \mu_0 I_{en}$

Where $\=B$ is the magnetic filed

∮ means the close integral

$d\= s$ is the length element

$\mu_0$ magnetic of permeability of free space

$I_{en}$ is the enclosed current

Now choosing a current path on a particular side of the solenoid (as shown on the second uploaded image) to evaluate the magnetic field

Where $I$ is the current elements within the enclosed current path

$N$ is the number of current elements in that particular enclosed current path.

and L is the length of enclosed path in the direction of the magnetic field as shown on the second uploaded image

So the integral of total path ∮ $\= B \cdot d \=s$ = $\int\limits { \= B } \, d\=s_1 + \int\limits { \= B } \, d\=s_2 +\int\limits { \= B } \, d\=s_3 +\int\limits { \= B } \, d\=s_4$

Where $s_1 ,s_2 ,s_3,s_4$ are the length element of each sides of the enclosed path

Now looking at the first integral we see that $s_1$ is moving the same direction with magnetic field B as shown on the third uploaded image

Now for $s_2$ the direction is perpendicular to the the direction of the magnetic field B so the value of magnetic field for that segment is 0 \

Now for $s_3$ from the third uploaded image we see that there are no this segment so the value is zero

$s_4$ segment is the same as $s_2$ segment so the value of magnetic field on this segment is zero

Therefore

∮ $\= B \cdot d \=s$ = $BL + 0 + 0 + 0$

Now from Ampere's law ∮ $\= B \cdot d \=s = \mu_0 I_{en}$ $\equiv$ $BL = \mu_0 I_{en}$

Now from the second uploaded image we see that $I_{en}$ would now be equal to $NI$

Therefore the equation becomes

$BL = \mu_0 (NI)$

Now making B the subject we have

$B = \frac{\mu_o NI}{L} = \mu_0 \frac{N}{L} I$

Now $\frac{N}{L} = n$ where n is the number of turns per length

So we have

$B = \mu_0 n I$

So the magnetic field on the solenoid 1 is would be

$B_1 (t) = \mu_0 n_1 I_1 (t)$

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The horse on a carousel is 3.5m from the central axis.A. If the carousel rotates at 0.13 rev/s , how long does it take the horse

Kitty [74]

Answer:

a. 15.4 seconds

b. 0.455 m/s

Explanation:

a. The carousel rotates at 0.13 rev/s.

This means that it takes the carousel 1 sec to make 0.13 of an entire revolution.

This means that time it will take to make a complete revolution is:

1 / 0.13 = 7.7 seconds

Therefore, the time it will take to make 2 revolutions is:

2 * 7.7 = 15.4 seconds

b. Let us calculate the linear velocity. Angular velocity is given as:

$\omega = v / r$

where v = linear velocity and r = radius

The radius of the circle is 3.5 m and the angular velocity is 0.13 rev/s, therefore:

0.13 = v / 3.5

v = 3.5 * 0.13 = 0.455 m/s

Linear velocity is 0.455 m/s

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3 years ago

You are given aqueous solutions of six different substances and asked to determine whether they are strong, weak, or nonelectrol

kogti [31]

Answer:

Answer is explained below;

Explanation:

Electrolytes are any substances that dissociate into charged particles called ions when dissolved in water. The positively charged ions called cations and the negatively charged ions called anions move toward the negative and positive terminals (cathode and anode) of an electric circuit.

When a substance dissolved in water completely dissociates into ions, it is called a strong electrolyte. The aqueous solutions containing strong electrolytes conduct electricity very well and the examples include strong acids and soluble ionic compounds such as barium chloride, sodium hydroxide, etc.

When a substance dissolved in water does not completely dissociate into ions, it is called a weak electrolyte. Since the aqueous solutions containing weak electrolytes have relatively few ions, their electrical conductivity is very low compared to the solutions containing strong electrolytes. Examples of weak electrolytes include weak acids and bases like acetic acid, ammonia, etc.

When a substance does not dissociate into ions when dissolved in water, it is called a nonelectrolyte. Since the aqueous solutions containing nonelectrolytes do not contain any ions, such solutions do not conduct electricity. Examples of nonelectrolytes are ethanol, aldehydes, glucose, ketones, etc.

If a solution contains dissolved ions, it conducts electricity and as the ion concentration increases, the conductivity also increases. To determine whether the aqueous solutions of six different substances are strong, weak, or nonelectrolytes, we can test them by applying a voltage to electrodes immersed in the solutions and a light bulb. By observing the brightness of the light bulb or by measuring the flow of electrical current, we can find out which solution contains a strong electrolyte or weak electrolyte, or nonelectrolyte.

If the solution contains a nonelectrolyte, the current flow is nil and the light bulb does not glow. If the solution contains a strong electrolyte, the current flow is very strong and so the brightness of the light bulb is very high. If the solution contains a weak electrolyte, the current flow is much low compared to the strong electrolyte and the light bulb glows, but the brightness is very low.

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3 years ago

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

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3 years ago

Learning Goal:

enot [183]

Answer:

A. $U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

$C=\dfrac{\epsilon A}{d}$

$C$ is the capacitance, $A$ is the common plate area, $d$ is the plate separation and $\epsilon$ is the permittivity of the material between the plates.

For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,

$U = \dfrac{CV^2}{2}$

A. Substituting for $C$ in $U$ ,

$U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. When the distance is $3d$ ,

$U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}$

$U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. When the distance is restored but with a dielectric material of dielectric constant, $K$ , inserted, we have

$U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

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2 years ago

Which of the following is a false statement about dispersion forces? View Available Hint(s) Which of the following is a false st

icang [17]

Explanation :

Dispersion forces are also known as London dispersion forces. It is the weakest force. Also, it is the part of the Van der Waals forces.

(1) This force is exhibited by all atoms and molecules.

(2) These forces are the result of the fluctuations in the electron distribution within molecules or atoms. Due to these fluctuations, the electric field is created. The magnitude of this force is explained in terms of Hamaker constant 'A'.

(3) Dispersion forces result from the formation of instantaneous dipoles in a molecule or atom. When electrons are more concentrated in a place, instantaneous dipoles formed.

(4) Dispersion force magnitude depends on the amount of surface area available for interactions. If the area increases, the size of the atom also increase. As a result, stronger dispersion forces.

So, the false statement is "Dispersion forces always have a greater magnitude in molecules with a greater molar mass".

4 0

3 years ago