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o-na [289]
2 years ago
3

A toroidal solenoid has a central radius of 0.50 m and a cross-sectional diameter of 10 cm. When a current passes through the co

il of the solenoid, the magnetic field inside the solenoid at its CENTER has a magnitude of 2.0 μT. What is the largest value of the magnetic field inside the solenoid when this current is flowing? (μ0 = 4π × 10^−7 T • m/A)
a. 1.8 micro T
b. 0.50 micro T
c. 2.8 micro T
d. 2.2 micro T
e. 3.5 micro T
Physics
1 answer:
sergejj [24]2 years ago
4 0

To solve this problem we will apply the concepts related to the magnetic field. From the magnetic field we will compare what happens with the distance or the radius between the bodies and we will make the replacement with the values we have. The largest value of magnetic field,

B = \frac{\mu_0 NI}{2\pi R}

Here,

\mu_0 = Permeability at free space

N = Number of loops

I = Current

R = Radius

From this relation we can conclude that the radius is inversely proportional to the Magnetic field, then

B \propto \frac{1}{R}

If we let the another values as constant we have that the relation between two magnetic field is,

\frac{B_1}{B_2} = \frac{R_2}{R_1}

B_2 = \frac{B_1}{R_2} R_1

Replacing,

B_2 = \frac{(2*10^{-6}T)}{(0.5m-0.05m)}(0.5m)

B_2 = 2.2\mu T

Therefore the correct option is 2.2\mu T

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Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli
Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: sin \theta=\frac{m\lambda}{\alpha}

Given m=3,\lambda=6.33\times 10^-^7 and \alpha=0.00015:

#Substitute our variable values in the minima equation to obtain \theta:

\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad

#draw a triangle to find the relationship between \theta, y \ and L.

tan(\theta)=y/L               #where y=4.05cm

L=y/tan(\theta)=3.20

Hence the screen is 3.20m from the split.

b.  To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

sin \theta=\frac{m\lambda}{\alpha}, where m=4 given that it's the minima with the smallest distance.

sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad

#we then use tan(\theta)=y/L to calculate L=4.5cm

Then from the equation subtract y_3 from y:

4.50cm-4.05cm=0.45cm

Hence, the distance \bigtriangleup y is 0.45cm

8 0
2 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
2 years ago
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H
kirill [66]

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

5 0
2 years ago
As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward
juin [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is   v =0.333 \  m/s in positive x -direction

The speed is s = 0.733 \ m/s

Explanation:

From the question we are told that

The distance from the house to truck is  D =  20 m

  The distance traveled back to retrieve  wind-blown hat is  d =  15

  The distance from the wind-blown hat position too the truck is  k =  20  m

  The total time taken is  t  =  75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

      L  =  20 - 15 + 20

=>    L  =  25 \ m

Generally the average velocity is mathematically represented as

          v  =  \frac{L}{t}

=>      v = \frac{25}{75}

=>      v =0.333 \  m/s

Generally the distance covered by Justin is mathematically represented as  

         R =  D+ d + k

=>      R =  20 + 15 +20

=>     R =  55 \  m

Generally Justin's average speed over a 75 s period is mathematically represented as

            s = \frac{R}{ t}

=>         s = \frac{55}{ 75}

=>        s = 0.733 \ m/s

8 0
2 years ago
Explain why it takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron fr
slava [35]
It takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron from a carbon atom because its inner core e, not valence e. C's 4th removed e is still a valence e. And also <span>because more nuclear charge acting on the second electron, it is more close to the nucleus, thus the the protons attract it more than the 4th electron.</span>
8 0
2 years ago
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