Question

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)

Answer 1

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

Answer 2

The magnitude of the magnetic field for this hollow cylinder is $2.1 \times 10^{-5}\;Tesla$

Given the following data:

Inner radius = 4 mm to m = 0.004 m.

Outer radius = 25 mm to m = 0.025 m.

Current = 4 A.

Distance = 17 mm to m = 0.017 m.

Permittivity of free space = $4\pi \times 10^{-7}\; T.m/A$

How to calculate the magnetic field.

Mathematically, the magnitude of the magnetic field for a hollow cylinder is given by this formula:

$B=\frac{\mu_o I(d^2 -r^2)}{2\pi d (R^2 -r^2)}$

Where:

• B is the magnetic field.

• I is the current.

• R is the outer radius.

• r is the inner radius.

• d is the distance.

• $\mu_o$ is the permittivity of free space.

Substituting the given parameters into the formula, we have;

$B=\frac{4\pi \times 10^{-7}\times 4 (0.017^2 -0.004^2)}{2\pi \times 0.017 (0.025^2 -0.004^2)}\\\\B=2.1 \times 10^{-5}\;Tesla$

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