Answer :
The electric field strength inside the silver wire is 3.398 × 10⁻⁸ V/m.
To find the electric field strength inside the wire, we can use the equation:
J = σE
where J is the current density (current per unit area), σ is the conductivity of the material, and E is the electric field strength.
First, we need to find the current density. The current is given as 36.9 mA, and the cross-sectional area of the wire can be found using the formula for the area of a circle:
A = πr²
where r is the radius of the wire, which is half of the diameter. So,
r = 4.71 mm / 2 = 2.355 mm = 0.002355 m
A = π(0.002355 m)² = 1.737 × 10⁻⁵ m²
The current density is then:
J = I / A = 36.9 × 10⁻³ A / 1.737 × 10⁻⁵ m² = 2.124 A/m²
Now we can plug in the values for J and σ to solve for E:
E = J / σ = 2.124 A/m² / 6.25 × 10⁷ Ω⁻¹ m⁻¹ = 3.398 × 10⁻⁸ V/m
You can learn more about electric fields at: brainly.com/question/15800304
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