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faust18 [17]
2 years ago
6

A long wire carrying a 5.8 A current perpendicular to the xy-plane intersects the x-axis at x=−2.3cm. A second, parallel wire ca

rrying a 3.0 A current intersects the x-axis at x=+2.3cm.
Required:
a. At what point on the x-axis is the magnetic field zero if the two currents are in the same direction?
b. At what point on the x-axis is the magnetic field zero if the two currents are in opposite directions?
Physics
1 answer:
adoni [48]2 years ago
8 0

Answer:

a) v    r = 0.7318 cm , b)  r = 7.23 cm

Explanation:

The magnetic field generated by a wire carrying a current can be found with Ampere's law

       ∫ B. ds = μ₀ I

the length of a surface circulates around the wire is

    s = 2π r

where r is the point of interest of the calculation of the magnetic field

         B = μ₀ I / 2π r

In this exercise we have two wires, write the equation of the magnetic field of each one

wire 1     I = 5.8 A

         B₁ = μ₀ 5,8 / 2π r₁

wire 2    I = 3.0 A

         B₂ = μ₀ 3/2π r₂

the direction of the field is given by the rule of the right hand, the thumb indicates the direction of the current and the other fingers the direction of the magnetic field

Let's apply these expressions to our case

a) the two streams go in the same direction

     using the right hand rule for each wire we see that between the two wires the magnetic fields have opposite directions so there is some point where the total value is zero

          B₁ - B₂ = 0

           B₁ = B₂

         μ₀ 5,8 / 2π r₁ = μ₀ 3 / 2π r₂

          5.8 / r₁ = 3 / r₂

          5.8 r₂ = 3r₁

the value of r is measured from each wire, therefore

        r₁ = 2.3 + r

        r₂ = 2.3 -r

we substitute

          5.8 (2.3 - r) = 3 (2.3 + r)

           r (3 + 5.8) = 2.3 (5.8 - 3)

           r = 2.3 2.8 / 8.8

           r = 0.7318 cm

b) the two currents have directional opposite

with the right hand rule in the field you have opposite directions outside the wires

suppose it is zero on the right side where the wire with the lowest current is

         B₁ = B₂

        5.8 / r₁ = 3 / r₂

        5.8 r₂ = 3 r₁

         r₁ = 2.3 + r

         r₂ = r - 2.3

        5.8 (r - 2.3) = 3 (2.3 + r)

        r (5.8 -3) = 2.3 (3 + 5.8)

        r = 2.3 8.8 / 2.8

        r = 7.23 cm

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In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because  the night -time temperature on grass and the car went below the dew point, but  the temperature of the concrete did not drop enough to reach the dew point  level

Therefore the correct option is therefore A

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i = i_1+i_2+i_3

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Question 24 1 pts Find the voltage in an extension cord having a 0.0600 22 resistance and through which a 5.00A current is flowi
Neporo4naja [7]

The voltage in the extension cord is 30 V.

The problem above can be solved using ohm's law

⇒ Formula:

V = IR.................. Equation 1

⇒ Where:

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  • R = Resistance of the extension cord.

From the question, I think there was a slight error in the value of the current given it suppose to be 500 A, and not 5.00 A

⇒ Given:

  • I = 500 A
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⇒ Substitute these values into equation 1

  • V = 500(0.06)
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Hence the voltage in the extension cord is 30 V

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Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

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when the frequency is doubled

\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

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when the frequency is doubled;

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Thus, the period will be reduced to one-half of what it was.

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I Know B for sure trust me had this quiz
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