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Step-by-Step Solution
Step 1: Identify the known quantities
• Radius of the cylindrical wire, R = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}
• Conductivity of the material, \sigma = 5 \times 10^{7} \ \text{S/m}
• Electric field applied, E = 10 \ \text{mV/m} = 10 \times 10^{-3} \ \text{V/m} = 0.01 \ \text{V/m}
Step 2: Write the formula for current density
The current density J in a conductor is given by
J = \sigma \, E.
Step 3: Calculate the current density
Substitute \sigma and E :
J = \left(5 \times 10^{7}\right) \times \left(0.01\right)
= 5 \times 10^{5} \ \text{A/m}^{2}.
Step 4: Relate current density to total current
Current I is the current density J multiplied by the cross-sectional area A of the wire. Since the wire is cylindrical:
A = \pi R^{2}.
Thus,
I = J \times A = J \times \pi R^{2}.
Step 5: Compute the current
Substitute J = 5 \times 10^{5} \ \text{A/m}^{2} and R = 0.5 \times 10^{-3} \ \text{m} :
I = \left(5 \times 10^{5}\right)\,\pi \,\left(0.5 \times 10^{-3}\right)^{2}.
First, compute \left(0.5 \times 10^{-3}\right)^{2} :
(0.5 \times 10^{-3})^{2} = 0.25 \times 10^{-6}.
Hence,
I = 5 \times 10^{5} \times \pi \times 0.25 \times 10^{-6}.
Combine the numerical factors:
I = \pi \times \left(5 \times 0.25\right) \times 10^{5 - 6}
= \pi \times 1.25 \times 10^{-1}
= 0.125\,\pi \ \text{A}.
Converting this to milliamperes (mA):
I = 0.125 \,\pi \ \text{A} = 125 \times 10^{-3}\,\pi \ \text{A} = 125\,\pi \ \text{mA}.
Step 6: Determine the value of x
According to the problem statement, the current in the wire is of the form x^{3} \pi \ \text{mA} . We have found:
I = 125\,\pi \ \text{mA}.
Thus,
x^{3} = 125 \quad \Longrightarrow \quad x = 5.
Final Answer
\boxed{5}
