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marusya05 [52]
3 years ago
9

The magnetic field produced by a long straight current-carrying wire is

Physics
1 answer:
alexdok [17]3 years ago
8 0

Answer:

proportional to the current in the wire and inversely proportional to the distance from the wire.

Explanation:

The magnetic field produced by a long, straight current-carrying wire is given by:

B=\frac{\mu_0 I}{2 \pi r}

where

\mu_0 is the vacuum permeability

I is the current intensity in the wire

r is the distance from the wire

From the formula, we notice that:

- The magnitude of the magnetic field is directly proportional to I, the current

- The magnitude of the magnetic field is inversely proportional to the distance from the wire, r

Therefore, correct option is

proportional to the current in the wire and inversely proportional to the distance from the wire.

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Answer

given,

Distance for decibel reading

 r₁ = 13 m

 r₂ = 24 m

When the engineer is at r₁ reading is β₁ = 101 dB

now, Calculating the Intensity at r₁

Using formula

\beta = 10 log(\dfrac{I}{I_0})

I₀ = 10⁻¹² W/m²

now inserting the given values

101= 10 log(\dfrac{I}{10^{-12}})

10.1= log(\dfrac{I}{10^{-12}})

\dfrac{I}{10^{-12}}= 10^{10.1}

I =10^{10.1} \times 10^{-12}

I = 0.01258 W/m²

now, calculating power at r₁

P₁ = I₁ A₁

P₁ = 0.01258 x 4 π r²

P₁ = 0.01258 x 4 π x 13²

P₁ = 26.72 W

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3 years ago
A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th
gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

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Length of the barbell = 1.90 m

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Mass on the right side = 37 kg

At the balance point: m_1x_1=m_2x_2

25 kg\times x=37 kg\times (1.90-x)

x=1.1338 m

The center of gravity is 1.1338 m away from the left side of the barbell

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