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zepelin [54]
4 years ago
15

A hollow cylinder with an inner radius of 4 mm and an outer radius of 25 mm conducts a 4-A current flowing parallel to the axis

of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 17 mm from its center? (μ 0 = 4π × 10-7 T · m/A)
Physics
1 answer:
ivann1987 [24]4 years ago
5 0

Answer:

2.1 × 10⁻⁵ T

Explanation:

Given:

Inner radius, r = 4 mm = 0.004 m

Outer radius, R = 25 mm = 0.025 m

Current, I = 4 A

Distance of the point from the center, a = 17 mm = 0.017 m

μ₀ = 4π × 10⁻⁷ T·m/A

Now,

For the hollow cylinder magnetic field (B) is given as:

B=\frac{\mu_0 I(a^2-r^2)\textup{}}{2\pi a(R^2-r^2)\textup{}}

on substituting the respective values, we get

B=\frac{4\pi\times10^{-7}\times4(0.017^2-0.004^2)\textup{}}{2\pi\times0.017(0.025^2-0.004^2)\textup{}}

or

B = 2.1 × 10⁻⁵ T

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Answer:

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Explanation:

a)

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b)

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final temperature of house, T_f=5.3+273=278.3\ K

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\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

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Answer:

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Explanation:

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Answer:

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