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3 years ago

If two wires run parallel and the current passes through both wires in the same direction, which happens to the wires?

2 answers:

5 0

I think the correct answer from the choices listed above is option B. If two wires run parallel and the current passes through both wires in the same direction, <span> the wires move apart because the magnetic fields of each wire are opposite, so they repel each other.</span>

skelet666 [1.2K]3 years ago

5 0

Answer:

A.) The wires move together because the magnetic field of each wire attracts the other wire.

Explanation:

As we know that the current in two wires are along same direction

so here magnetic field due to one wire on the adjacent wire will produce magnetic force

this magnetic force is given as

$F = i(\vec L \times \vec B)$

since the direction of length vector is along the current in the wire so here the force on two wires due to magnetic field of its adjacent wire is always towards each other.

So two current carrying wires will attract each other when current flows in the two wires in same direction

so correct answer will be

A.) The wires move together because the magnetic field of each wire attracts the other wire.

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A spring with a constant of 92N/m is compressed 2.8 cm. How much potential energy is stored in the spring?

Arada [10]

Answer:

0.036J

Explanation:

Given parameters:

Spring constant , K = 92N/m

Compression = 2.8cm = 0.028m

Unknown:

Potential energy = ?

Solution:

To solve this problem;

P.E = $\frac{1}{2}$ K e²

K is the spring constant

e is the compression

so;

P.E = $\frac{1}{2}$ x 92 x 0.028² = 0.036J

6 0

3 years ago

A traveling electromagnetic wave in a vacuum has an electric field amplitude of 50.9 V/m. Calculate the intensity ???? of this w

Sliva [168]

Answer:

3.44 W/m²

1.134 J

Explanation:

E₀ = Intensity of electric field = 50.9 V/m

I = Intensity of electromagnetic wave

Intensity of electromagnetic wave is given as

I = (0.5) ε₀ E₀² c

I = (0.5) (8.85 x 10⁻¹²) (50.9)² (3 x 10⁸)

I = 3.44 W/m²

A = Area = 0.0277 m²

t = time interval = 11.9 s

Amount of energy is given as

U = I A t

U = (3.44) (0.0277) (11.9)

U = 1.134 J

5 0

3 years ago

Using the result of the preceding problem, (a) calculate the distance between fringes for 633-nm light falling on double slits s

vekshin1

To solve this problem it is necessary to apply the concepts related to the concept of superposition and the fringe separation for double slit experiment.

The equation can be written as

$\Delta y = \frac{x\lambda}{d}$

Where

$\Delta y$ = Distance between fringes

x = distance between slits and screen

d = Distance between slits

$\lambda$ = Wavelength

Our values are given as

d= 0.08mm

x =3m

$\lambda = 633nm$

In this way replacing in the equation,

$\Delta y = \frac{x\lambda}{d}$

$\Delta y = \frac{3m(633nm*(\frac{1*10^{-9}}{1nm}))}{0.08*(\frac{1*10^{-3}m}{1mm})}$

$\Delta y = 2.37*10^{-2}m$

$\Delta y = 2.37cm$

Therefore the distance between the fringes is 2.37cm

PART B) For the case in which it is submerged in water it is necessary to apply the relationship of the fringes with the index of refraction therefore

$\Delta Y_2 = \frac{\Delta Y_1}{n}$

$\Delta Y_2 = \frac{2.37}{1.33}$

$\Delta Y_2 = 1.78cm$

3 0

4 years ago

What is the electric field strength just outside the flat surface of the conductor?

inna [77]

We can find the answer step-by-step:

1) The electric charges on a conductor must lie entirely on its surface. This is because the charges have same sign, so the force acting between each other is repulsive therefore the charges must be as far apart as possible, i.e. on the surface of the conductor.

2) We consider a cylinder perpendicular to the surface of the conductor, that crosses the surface with its section. We then apply Gauss law, which states that the flux of the electric field through this cylinder is equal to the total charge inside it divided the electrical permittivity:
$\Phi = \frac{Q}{\epsilon_0}$

3) The electric field outside the surface is perpendicular to the surface itself (otherwise there would be a component of the electric force parallel to the surface, which would move the charge, violating the condition of equilibrium). The electric field inside the conductor is instead zero, because otherwise charges would move violating again equilibrium condition. Therefore, the only flux is the one crossing the section A of the cylinder outside the surface:
$\Phi = E A$

4) The total charge contained in the cylinder is the product between the section, A, and the charge density $\sigma$ on the surface of the conductor:
$Q=\sigma A$

5) Substituting the flux and the charge density inside Gauss law, we can find the electric field just outside the surface of the conductor:
$EA= \frac{\sigma A}{\epsilon_0}$
therefore
$E= \frac{\sigma}{\epsilon_0}$

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4 years ago

Scientific experiments are specially designed to test explanations for an observed phenomenon. A student decides to conduct an e

makkiz [27]

The amount of water given is a control. D) because, if a plant grows bigger as a result of the water, and not the sun this would disrupt the experiment.

8 0

3 years ago

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