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Luda [366]
3 years ago
11

Two points, point A and point B, are situated above a current-carrying wire. Point B is located at a distance R from the wire, w

hich is twice as far from the wire as point A. By what factor is the magnetic field at point A larger or smaller than the magnetic field at point B?
Physics
1 answer:
padilas [110]3 years ago
7 0

Explanation:

It is given that, there are two points situated above a current-carrying wire. Point B is located at a distance R from the wire, which is twice as far from the wire as point A such that,

r_B=R

And r_A=R/2

We know that the magnetic field at a distance R is inversely proportional to the distance from wire as :

B=\dfrac{\mu_o}{2\pi}\dfrac{I}{r}

B_A\propto\dfrac{1}{(R/2)}

B_B\propto\dfrac{1}{(R)}

\dfrac{B_A}{B_B}=\dfrac{1/(R/2)}{1/R}

\dfrac{B_A}{B_B}=\dfrac{2}{1}

B_A=2B_B

So, the magnetic field at point A larger than the magnetic field at point B by a factor of 2. Hence, this is the required solution.

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A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina
Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy U=4.30\times10^{3}\ J

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

\Delta S=\dfrac{\Delta U}{T}

Where, \Delta U = internal energy

T = average temperature

Put the value in to the formula

\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}

\Delta S=-14.72\ J/K

Hence, The approximate change in entropy is -14.72 J/K.

5 0
3 years ago
A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.
Pavlova-9 [17]

Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

 

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

         v₂ = \sqrt {2g \ y_1}

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

6 0
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Lemur [1.5K]

Explanation:

Work cannot be increased by using a machine of some kind.

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The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif
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The relation between potential difference and the electric field is given by ΔV = E.d

Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.

The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.

The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.

Refer to more about the potential difference here

brainly.com/question/12198573

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strojnjashka [21]

Answer:

C

Explanation:

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