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kirill [66]
3 years ago
13

You are given two infinite, parallel wires, each carrying current.The wires are separated by a distance, and the current in the

two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
A. Is the force between the wires attractive orrepulsive?
B. What is the force per unit length between the two wires?
Physics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

A. Attractive

B. ( μ₀I² ) / ( 2πd )

Explanation:

A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.

B. Suppose that two wires of length l_1 and l_2 both carry the current I in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -

B_1 = μ₀I / 2πd

The force experienced by wire 2 should thus be -

F_2 = I( l_2 * B_1 )

= I * l_2 * B_1 * Sin( 90 )

= I * l_2 ( μ₀I / 2πd )

Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...

( F_2 / l_2 ) = ( μ₀I² ) / ( 2πd ) ... which is our solution

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A cube has a density of 1800 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experimenter in
olga2289 [7]

Answer:

4341.44763 kg/m³

Explanation:

\rho' = Actual density of cube = 1800 kg/m³

\rho = Density change due to motion

v = Velocity of cube = 0.91c

c = Speed of light = 3\times 10^8\ m/s

Relativistic density is given by

\rho=\frac{\rho'}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-\frac{0.91^2c^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-0.91^2}}\\\Rightarrow \rho=4341.44763\ kg/m^3

The cube's density as measured by an experimenter in the laboratory is 4341.44763 kg/m³

5 0
4 years ago
The magnetic flux through a coil of wire containing two loops changes at a constant rate from -58 wb to 38 wb in 0. 34 s. What i
german

The emf induced in the coil is -5.65 V

<h3>Induced emf in coil</h3>

The induced emf in the coil is given by ε = -NΔΦ/Δt where

  • ΔΦ = change in magnetic flux Φ₂ - Φ₁ where
  • Φ₁ = initial magnetic flux = -58 Wb and
  • Φ₂ = final magnetic flux = 38 Wb and and
  • Δt = change in time = t₂ - t₁ where
  • t₁ = initial time = 0 s and
  • t₂ = final time = 34 sand
  • N = number of loops of coil = 2

Since ε = -NΔΦ/Δt

ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)

Substituting the values of the variables into the equation, we have

ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)

ε = -2(38 Wb - (-58 Wb))/(34 s - 0 s)

ε = -2(38 Wb + 58 Wb)/(34 s - 0 s)

ε = -2(96 Wb)/34 s

ε = -192 Wb/34 s

ε = -5.65 Wb/s

ε = -5.65 V

So, the emf induced in the coil is -5.65 V

Learn more about induced emf in coil here:

brainly.com/question/13051297

6 0
3 years ago
A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if
tigry1 [53]

Answer:

4.16m/s²

Explanation:

According to Newtons second law;

\sum Fx = ma_x\\Fm - Ff = ma_x\\W sin\theta - \mu R cos \theta = ma_x\\mg sin\theta - \mu mg cos\theta = ma_x\\

Fm is the moving force

\mu is  the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is  4.16m/s²

4 0
3 years ago
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