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3 years ago

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1. A 2,000-turn solenoid is 65 cm long and has cross-sectional area 30 cm2. What rate of change of current will produce a 600 Vo

lts emf in this solenoid.

1 answer:

JulsSmile [24]3 years ago

5 0

Answer:

$\frac{dI}{dt} = 2.59\ x\ 10^4\ A/s$

Explanation:

First, we will calculate the inductance of the solenoid by using the following formula:

$L = \frac{\mu_o AN^2}{l}$

where,

L = self-inductance of solenoid = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

A = Cross-sectional area = 30 cm² = 3 x 10⁻³ m²

N = No. of turns = 2000

l = length = 65 cm = 0.65 m

Therefore,

$L = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(3\ x\ 10^{-3}\ m^2)(2000)^2}{0.65\ m}\\\\L = 0.0232\ H$

Now, we will use Faraday's law to calculate the rate of change of current:

$emf = L\frac{dI}{dt}\\\\ \frac{dI}{dt} =\frac{emf}{L} \\\\ \frac{dI}{dt} =\frac{600\ V}{0.0232\ H}\\\\ \frac{dI}{dt} = 2.59\ x\ 10^4\ A/s$

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Answer:

Part a)

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Part b)

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Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

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here we will have

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now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

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Answer:

50 Mph.

Explanation:

According to the National Severe Storms Laboratory, winds can really begin to cause damage when they reach <em><u>50 mph</u></em>. But here’s what happens before and after they reach that threshold, according to the Beaufort Wind Scale (showing estimated wind speeds): - at 19 to 24 mph, smaller trees begin to sway.

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Given that: = 2i + 9j +3k and = -i – 4k . Find

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3i+9j= -7k

Explanation:

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2 years ago

A ship coming from sea enters the river ,with its hull sink more or less in river water?Give reasons.

GenaCL600 [577]

Explanation:

Hey, there!

Answer: It sinks more.

We know that the water of sea is salty in nature which increases the density of water in sea and donot let's ship sink but in condition of river water it doesn't have any effects of salty water as it doesn't contain salty water due to which it's density is no more than the densityof ship which let's it sink more than sea water.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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3 years ago

A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .

As a result,

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V$ .

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3 years ago