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alex41 [277]
3 years ago
10

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

×10−2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 6.40×10−5 T . Part A
What is the total magnitude of the magnetic flux ( ?initial) through the coil before it is rotated?

Express your answer numerically, in webers, to at least three significant figures.

Part B

What is the magnitude of the total magnetic flux ?final through the coil after it is rotated?

Express your answer numerically, in webers, to at least three significant figures.

Part C

What is the magnitude of the average emf induced in the coil?

Express your answer numerically (in volts) to at least three significant figures.
Physics
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

  • no. of turns i the coil, n=200
  • area of the coil, a=13.1 \times 10^{-4}\,m^2
  • time interval of rotation, t=3.1\times 10^{-2}\,s
  • intensity of magnetic field, B=6.4\times 10^{-5}\,T

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

\Phi=83.84\times 10^{-9} Wb

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

\Phi=0 Wb

(C)

According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

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Calculate the molarity of a 10. 0% cacl₂ solution. The density of the solution is 1. 0835 g/cm³.
brilliants [131]

The molarity of 10% CaCl2 is 0.9%

concentration of the given salt CaCl₂ = 10%

Density of a solution = 1.0835 g/cm³

Volume = m / d

= 100 / 1.0835

= 92.29 litres

Density = mass / volume

1.0835 × 92.29 = mass

mass = 99.99 gram

Thus the molarity can be calculated by = moles of solute / volume of solution multiplied by 100

= 0.9008/ 92.29 X 100 %

= 0.009 X 100 %

= 0.9 %

The molarity of 10% CaCl2 is 0.9%

To know more about density and molarity you may visit the link which is mentioned below:

brainly.com/question/10710093

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5 0
1 year ago
a wave is described by where x is in meters, y is in centimeters and t is in seconds. The angular frequency is
Sergeeva-Olga [200]

Complete Question

A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave  frequency is

Answer:

The  value is w =  10 \ rad /s

Explanation:

From the question we are told that  

    The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)

Generally the sinusoidal equation representing the motion of a wave is mathematically represented as

         y(x,t) =  Asin(kx + wt )

Where  w  is the  angular frequency

Now comparing this equation  with that given we see that

       w =  10 \ rad /s

 

               

7 0
3 years ago
Does the earths magnetic field change with time?
ZanzabumX [31]

Answer:

As per the fossil fuel records, magnetic field reversal does not impact living beings. It will take almost a century for the poles to complete the shift. Meanwhile, the earth is left with almost zero magnetic field.

6 0
3 years ago
A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel
lana66690 [7]

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

\epsilon = - d\frac{\phi_B}{dt}

\epsilon = -NA\frac{dB}{dt}

Where,

N = Number of loops

A = Cross sectional Area

B = Magnetic Field

\epsilon = (15)(\pi(0.03)^2)\frac{0-0.5}{0.1}

\epsilon = 0.212V

\epsilon = 0.21V

Therefore the correct answer is A.

6 0
3 years ago
In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th
Radda [10]

Answer:

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

8 0
3 years ago
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