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3 years ago

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A hollow cylinder with an inner radius of and an outer radius of conducts a 3.0-A current flowing parallel to the axis of the cy

linder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point from its center?

1 answer:

Artemon [7]3 years ago

7 0

Complete Question:

A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-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 12mm from its center?

Answer:

The magnitude of the magnetic field = 7.24 μT

Explanation:

Inner radius, a = 4.0 mm = 0.004 m

Outer radius, b = 30 mm = 0.03 m

Radius, r = 12 mm = 0.012 m

let h² = b² - a²

h² = 0.03² - 0.004²

h² = 0.000884

Let d² = r² - a²

d² = 0.012² - 0.004²

d² = 0.000128

Current I = 3A

μ = 4π * 10⁻⁷

The magnitude of the magnetic field is given by:

$B = \frac{\mu I d^{2} }{2\pi r h^{2} } \\B = \frac{4\pi * 10^{-7} * 3* 0.000128^{2} }{2\pi *0.012* 0.000884^{2} }$

B = 7.24 * 10⁻⁶T

B = 7.24 μT

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