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Murrr4er [49]
3 years ago
14

By what factor is the self-inductance of an air solenoid changed if its length and number of coil turns are both tripled

Physics
1 answer:
fredd [130]3 years ago
4 0

Answer:

The new self inductance is 3 times of the initial self inductance.

Explanation:

The self inductance of a solenoid is given by :

L=\dfrac{\mu_oN^2 A}{L}

Where

N is number of turns per unit length

A is area of cross section

l is length of solenoid

If length and number of coil turns are both tripled,

l' = 3l and N' = 3N

New self inductance is given by :

L'=\dfrac{\mu_oN'^2 A}{L'}\\\\=\dfrac{\mu_o(3N)^2 A}{3L}\\\\=3\dfrac{\mu_oN^2 A}{L}\\\\=3L

So, the new self inductance is 3 times of the initial self inductance.

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For your final exam in electronics, you’re asked to build an LC circuit that oscillates at 10 kHz. In addition, the maximum curr
Marina CMI [18]

Answer:

 L= 2 mH

C=1.26\times 10^{-7}\ F            

Explanation:

Given that

Frequency , f= 10 kHz

Maximum current ,I = 0.1 A

Maximum energy stored ,E= 1 x 10⁻⁵ J

The maximum energy stored in the inductor is given as follows

E=\dfrac{1}{2}LI^2

Where ,L= Inductance

I=Current

E=Energy

Now by putting the values in the above equation

10^{-5}=\dfrac{1}{2}\times L\times 0.1^2

L=\dfrac{2\times 10^{-5}}{0.1^2}\ H

L=0.002 H

L= 2 mH

We know that frequency f is given as

2\pi f=\dfrac{1}{\sqrt{LC}}

C=Capacitance  , f=frequency ,L=Inductance

Now by putting the values

2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }

62831.85=\dfrac{1}{\sqrt{0.002\times C}}

\sqrt{0.002\times C=\dfrac{1}{62831.85}

0.002\times C=0.0000159^2

C=\dfrac{0.0000159^2}{0.002}\ F

C=1.26\times 10^{-7}\ F

Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.

6 0
3 years ago
A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
What is the substance that is broken down in cellular respiration
Harman [31]
Glucose.... have a nice day
4 0
3 years ago
When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil
Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

2t = 2\lambda_{film}

t = \lambda_{film}

The index of refraction of soap is given, then

\lambda_{film} = \frac{\lambda_{vacuum}}{n}

Combining the results of all steps we get

t = \frac{\lambda_{vacuum}}{n}

Rearranging, we find

\lambda_{vacuum} = tn

\lambda_{vacuum} = (278)(1.33)

\lambda_{vacuum} = 369.74nm

4 0
3 years ago
An inductor has inductance of 0.260 H and carries a current that is decreasing at a uniform rate of 18.0 mA/s.
nignag [31]

Answer:

The self-induced emf in this inductor is 4.68 mV.

Explanation:

The emf in the inductor is given by:

\epsilon = -L\frac{dI}{dt}

Where:

dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)

L: is the inductance = 0.260 H

So, the emf is:

\epsilon = -L\frac{dI}{dt} = -0.260 H*(-18.0 \cdot 10^{-3} A/s) = 4.68 \cdot 10^{-3} V

Therefore, the self-induced emf in this inductor is 4.68 mV.  

I hope it helps you!

6 0
3 years ago
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