Question

A 2.9 mm-diameter copper wire carries a 40 A current (uniform Part A across its cross section). Determine the magnetic field at the surface of the wire. Express your answer using two significant figures. Determine the magnetic field inside the wire, 0.50 mm below the surface. Express your answer using two significant figures. Determine the magnetic field outside the wire 2.5 mm from the surface. Express your answer using two significant figures.

Answer 1

The magnetic field at the surface of the wire is 0.069 T. The magnetic field inside the wire, 0.50 mm below the surface, is 0.064 T. The magnetic field outside the wire, 2.5 mm from the surface, is 0.041 T.

Step-by-step explanation:

To determine the magnetic field at the surface of the wire, we can use Ampere's law.

Ampere's law states that the magnetic field along a closed loop is equal to the product of the current passing through the loop and the permeability of free space divided by the circumference of the loop.

The magnetic field at the surface of the wire can be calculated using the formula:

B = μ0 * I / (2πr), where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the radius of the wire.

Substituting the given values:

B = (4πx10^-7 T*m/A) * 40 A / (2π0.0029 m)

B = 0.069 T

The magnetic field inside the wire, 0.50 mm below the surface, can be calculated using the same formula:

B = μ0 * I / (2πr), where r is the radius of the wire plus the distance below the surface.

Substituting the given values:

B = (4πx10^-7 T*m/A) * 40 A / (2π(0.0029 m + 0.0005 m))

B = 0.064 T

The magnetic field outside the wire, 2.5 mm from the surface, can also be calculated using the same formula:

B = μ0 * I / (2πr), where r is the radius of the wire plus the distance from the surface.

Substituting the given values:

B = (4πx10^-7 T*m/A) * 40 A / (2π(0.0029 m + 0.0025 m))

B = 0.041 T

Answer 2

The magnetic field at the surface of the wire is 0.055 T. The magnetic field inside the wire, 0.50 mm below the surface, is 0.223 T. The magnetic field outside the wire, 2.5 mm from the surface, is 0.025 T.

Step-by-step explanation:

To determine the magnetic field at the surface of the wire, we can use Ampere's Law. Ampere's Law states that the magnetic field at a distance r from a long straight wire carrying current I is given by B = μ₀I/2πr, where μ₀ is the permeability of free space. In this case, the diameter of the wire is 2.9 mm, which means its radius is 1.45 mm or 0.00145 m. Plugging in the values, we get:

B = (4π × 10^-7 T·m/A) × (40 A) / (2π × 0.00145 m) = 0.055 T

So, the magnetic field at the surface of the wire is 0.055 T.

To determine the magnetic field inside the wire, 0.50 mm below the surface, we can use the same formula, but with a smaller radius. The radius inside the wire will be 0.00145 m - 0.00050 m = 0.00095 m. Plugging in the values, we get:

B = (4π × 10^-7 T·m/A) × (40 A) / (2π × 0.00095 m) = 0.223 T

So, the magnetic field inside the wire, 0.50 mm below the surface, is 0.223 T.

To determine the magnetic field outside the wire, 2.5 mm from the surface, we can use the same formula, but with a larger radius. The radius outside the wire will be 0.00145 m + 0.00250 m = 0.00395 m. Plugging in the values, we get:

B = (4π × 10^-7 T·m/A) × (40 A) / (2π × 0.00395 m) = 0.025 T

So, the magnetic field outside the wire, 2.5 mm from the surface, is 0.025 T.