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kotegsom [21]
3 years ago
7

You are designing a new solenoid and experimenting with material for each turn. The particular turn you are working with is a ci

rcular loop of radius 3.00 cm that carries a current of 12.0 A . Calculate the magnetic field B at the center of the loop.
Physics
1 answer:
andriy [413]3 years ago
8 0

Answer:

The magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.

Explanation:

Given;

Radius of circular loop, R = 3.00 cm = 0.03 m

Current in the loop, I = 12.0 A

Magnetic field at the center of circular loop is given as;

B = μ₀I / 2R

Where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

R is the radius of the circular loop

I is the current in the loop

Substitute the given values in the above equation and calculate the magnitude of the magnetic field;

B = (4π x 10⁻⁷ x 12)/ 0.03

B = 5.0272 x 10⁻⁴ T

Therefore, the magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.

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1) Consider a source particle of charge qS=9 C located at (−9,5) [distances in meters]. Find the electric field vector at the ta
xeze [42]

Answer:

1)  E = 2.25 i^+ 0.809j^)  10⁹ N / C , 2)   E = 2.39 10⁹ N / C , 3)    θ = 19.8º , 4)   F = 19.12 10⁹ N , 5)  E = (1.32 i^+ 3.56 j^) 109 N/C

Explanation:

1) The equation for the electric field is

       E = k q / r²

Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point

Since the electric field is a vector magnitude, we can find its component

X axis

     Ex = k q / x²

where the distance on the axis is

      x = √ (X₂-x₁)²

      x = √ (-15 + 9)² = 6 m

      Eₓ = 8.99 10⁹ 9/6²

      Eₓ = 2.25 10⁹ N /C

Y Axis

     y = √ (y₂-y₁)² = √ (15-5)² = 10 m

     E_{y} = 8.99 10⁹ 9/10²

      E_{y}  = 0.809 10⁹ N / C

     

     E = Eₓ i^+   E_{y}  j^

     E = 2.25 i^+ 0.809j^)  10⁹ N / C

2) the magnitude can be found using the Pythagorean triangle

        E = √ (Eₓ² +  E_{y}²)

        E = √ (2.25² + 0.809²) 10⁹

        E = 2.39 10⁹ N / C

3) to find the angle let's use trigonometry

       tan θ = E_{y} / Eₓ

       θ = tan⁻¹ E_{y} / Eₓ

       θ = tan⁻¹ (0.809 / 2.25)

       θ = 19.8º

Regarding the positive side of the x axis

4) a charge  q2 = 8C is placed, let's calculate the force

      F = q E

      F = 8 2.39 10⁹

      F = 19.12 10⁹ N

5) The total electric field at the origin, let's look for its components

     q₁ = 9C

     r₁ = -9 i ^ + 5 j ^

     q₂ = 8 C

     r₂ = -15 i ^ + 15 j ^

X axis

     Eₓ = E₁ₓ + E₂ₓ

     Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²

     Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)

     Eₓ = 1.32 109 N / A

Y Axis

    E_{y} =  E_{1y} + E_{2y}

      E_{y} = k q₁ / Δy₁² + k q₂ / Δy₂²

      E_{y} = 8.99 109 (9/5² + 8/15²)

      E_{y} = 3.56 109 N / A

     E = (1.32 i^+ 3.56 j^) 109 N/C

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