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kirill [66]
3 years ago
13

A small current element carrying a current of I = 5.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

field, d → B , at the locations specified. Enter the correct magnitude and select the direction from the list. If the direction is negative, indicate this by entering the magnitude as a negative number. What is the magnitude and direction of d → B on the x ‑axis at x = 2.00 m ?
Physics
1 answer:
defon3 years ago
7 0

Answer:

dB = (-5 × 10⁻⁷k) T

Magnitude of dB = (-5 × 10⁻⁷) T; magnitude is actually (5 × 10⁻⁷) T in the mathematical sense of what magnitude is, but this question instructs to include the negative of the direction of dB is negative.

Direction is in the negative z-direction as evident from the sign on dB's vector notation.

Explanation:

From Biot Savart's relation, the magnetic field is given by

dB = (μ₀I/4πr³) (dL × r)

μ₀ = (4π × 10⁻⁷) H/m

I = 5.0 A

r = (2î) m

Magnitude of r = 2

dL = (4j)

(dL × r) is the vector product of both length of current carrying wire vector and the vector position of the point where magnetic field at that point is needed.

(dL × r) = (4j) × (2î)

​|i j k|

|0 4 0|

|2 0 0|

(dL × r) = (0î + 0j - 8k) = (-8k)

(μ₀I/4πr³) = (4π × 10⁻⁷ × 5)/(4π×2³)

(μ₀I/4πr³) = (6.25 × 10⁻⁸)

dB = (μ₀I/4πr³) (dL × r) = (6.25 × 10⁻⁸) × (-8k)

dB = (-5 × 10⁻⁷k) T

Magnitude of dB = (-5 × 10⁻⁷) T; magnitude is actually (5 × 10⁻⁷) T in the mathematical sense of what magnitude is, but this question instructs to include the negative of the direction of dB is negative.

Direction is in the negative z-direction as evident from the sign on dB's vector notation.

Hope this Helps!!!

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Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

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Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

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The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

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At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

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(a) The ideal gas equation is given as follows;

P·V = n·R·T

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∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

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T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

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The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

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Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

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∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

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