To solve this problem, we need to use Ampere's Law, which states that the integral of the magnetic field B around any closed loop is equal to μ0 times the net current enclosed by the loop.

The current density J in the wire is given by the total current I divided by the cross-sectional area A of the wire. In this case, the cross-sectional area of the wire is (0.15 m)^2 = 0.0225 m^2, and the total current is 15 A, so the current density is J = I/A = 15 A / 0.0225 m^2 = 666.67 A/m^2.

The current enclosed by the smaller square loop is the current density times the cross-sectional area of the smaller loop. The side length of the smaller loop is 0.06 m, so its cross-sectional area is (0.06 m)^2 = 0.0036 m^2. Therefore, the enclosed current is J * 0.0036 m^2 = 666.67 A/m^2 * 0.0036 m^2 = 2.4 A.

According to Ampere's Law, the integral of B around the loop is equal to μ0 times the enclosed current. The permeability of free space μ0 is 4π x 10^-7 T m/A, so the integral of B is μ0 * 2.4 A = 4π x 10^-7 T m/A * 2.4 A = 3.01 x 10^-6 T m.

This is the value of the integral ∮𝐵⋅𝑑𝑠 along the smaller square loop.

Question

To solve this problem, we need to use Ampere's Law, which states that the integral of the magnetic field B around any closed loop is equal to μ0 times the net current enclosed by the loop.

The current density J in the wire is given by the total current I divided by the cross-sectional area A of the wire. In this case, the cross-sectional area of the wire is (0.15 m)^2 = 0.0225 m^2, and the total current is 15 A, so the current density is J = I/A = 15 A / 0.0225 m^2 = 666.67 A/m^2.

The current enclosed by the smaller square loop is the current density times the cross-sectional area of the smaller loop. The side length of the smaller loop is 0.06 m, so its cross-sectional area is (0.06 m)^2 = 0.0036 m^2. Therefore, the enclosed current is J * 0.0036 m^2 = 666.67 A/m^2 * 0.0036 m^2 = 2.4 A.

According to Ampere's Law, the integral of B around the loop is equal to μ0 times the enclosed current. The permeability of free space μ0 is 4π x 10^-7 T m/A, so the integral of B is μ0 * 2.4 A = 4π x 10^-7 T m/A * 2.4 A = 3.01 x 10^-6 T m.

This is the value of the integral ∮𝐵⋅𝑑𝑠 along the smaller square loop.

Knowee AI · Accepted Answer

To solve this problem, we need to use Ampere's Law, which states that the integral of the magnetic field B around any closed loop is equal to μ0 times the net current enclosed by the loop.

The current density J in the wire is given by the total current I divided by the cross-sectional area A of the wire. In this case, the cross-sectional area of the wire is (0.15 m)^2 = 0.0225 m^2, and the total current is 15 A, so the current density is J = I/A = 15 A / 0.0225 m^2 = 666.67 A/m^2.

The current enclosed by the smaller square loop is the current density times the cross-sectional area of the smaller loop. The side length of the smaller loop is 0.06 m, so its cross-sectional area is (0.06 m)^2 = 0.0036 m^2. Therefore, the enclosed current is J * 0.0036 m^2 = 666.67 A/m^2 * 0.0036 m^2 = 2.4 A.

According to Ampere's Law, the integral of B around the loop is equal to μ0 times the enclosed current. The permeability of free space μ0 is 4π x 10^-7 T m/A, so the integral of B is μ0 * 2.4 A = 4π x 10^-7 T m/A * 2.4 A = 3.01 x 10^-6 T m.

This is the value of the integral ∮𝐵⋅𝑑𝑠 along the smaller square loop.

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