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vivado [14]
1 year ago
8

a long wire with a square cross-section 13 cm on a side carries a current of 10 a that is uniformly distributed over the cross-s

ection of the wire. what is the value of the integral along a square with side length 8 cm concentric with the wire and with sides parallel to those of the wire?
Physics
1 answer:
kicyunya [14]1 year ago
3 0

The value of the integral along a square with side length 8 cm concentric with the wire and with sides parallel to those of the wires 4.7*10^-6 I-m.

The integral around a closed path of the component of the magnetic field tangent to the direction of the path equals µ0 times the current intercepted by the area within the path.

§ B.ds=UoI

Thus the line integral (circulation) of the magnetic field around some arbitrary closed curve is proportional to the total current enclosed by that curve.

Ampere’s Law can be useful when calculating magnetic fields of current distributions with a high degree of symmetry (similar to symmetrical charge distributions in the case of Gauss’ Law)

The total magnetic circulation is zero only in the the enclosed net current is zero, the magnetic field is normal to the selected path at any point the magnetic field is zero.

According to the amperes law,

§ B.ds=UoI

I is the current through Amperian loop

So,here I= A² Io/Ao^2 (as current is uniformly distributed)

              = 8^2*10/13*13 = 3.79A

§ B.ds=UoI

         = 4π*10^-7*3.79

         = 4.7*10^-6 I-m

To know more about ampere's law here

brainly.com/question/17070619

#SPJ4

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What is the gravitational force between Earth and a 10 kg object on Earth’s surface?
4vir4ik [10]

Answer:

The force between the Earth and the object is, F = 9.61 x 10⁻¹² N

Explanation:

Given data,

The mass of the Earth, M = 5.9 x 10²⁴ kg

The mass of the object, m = 10 kg

The gravitational force between the earth and the object is given by the formula,

                         F = G Mm/R²

Substituting the given values in the above equation,

                         F = 6.673 x 10⁻¹¹ x 5.9 x 10²⁴ x 10 / 6.4 x 10⁶

                            =  9.61 x 10⁻¹² N

Hence, the force between the Earth and the object is, F = 9.61 x 10⁻¹² N

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6 0
3 years ago
consider two stars, star a and star b. star a has a temperature of 4900 k , and star b has a temperature of 9900 k . how many ti
ValentinkaMS [17]

The Energy flux from Star B is 16 times of the energy flux from Star A.

We have Two stars - A and B with 4900 k and 9900 k surface temperatures.

We have to determine how many times larger is the energy flux from Star B compared to the energy flux from Star A.

<h3>State Stephen's Law?</h3>

Stephens law states that if E is the energy radiated away from the star in the form of electromagnetic radiation, T is the surface temperature of the star, and σ is a constant known as the Stephan-Boltzmann constant then-

$\frac{Energy}{Area} = \sigma\times T^{4}

Now -

Energy emitted per unit surface area of Star is called Energy flux. Let us denote it by E. Then -

$E= \sigma\times T^{4}

Now -

For Star A →

T_{A} = 4900 K

For Star B →

T_{B} = 9900 K

Therefore -

$\frac{T_{B} }{T_{A} } =\frac{9900}{4900}

\frac{T_{B} }{T_{A} }= 2.02 = 2 (Approx.)

Now -

Assume that the energy flux of Star A is E(A) and that of Star B is E(B). Then -

$\frac{E(B)}{E(A)} = \frac{\sigma\times T(B)^{4} }{\sigma \times T(A)^{2} }

E(B) = E(A) x (\frac{T(B)}{T(A)} )^{4}

E(B) = E(A) x 2^{4}

E(B) = 16 E(A)

Hence, the Energy flux from Star B is 16 times of the energy flux from Star A.

To learn more about Stars, visit the link below-

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4 0
2 years ago
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

However, on the disk, moment of inertIa is:

I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

Frictional torques by the wall on the disk is:

T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

F= \dfrac{0.6}{0.2} \\ \\ F = 3N

8 0
3 years ago
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