Question
Download Solution PDFA copper wire of length 10 m and radius \(\left(10^{-2} / \sqrt{\pi}\right) \) m has electrical resistance of 10Ω . The current density in the wire for an electric field strength of 10 (V/m) is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
The current density is defined as the product of conductivity and electric field and it is written as;
J = σE
Here, σ is the conductivity and E is the electric field.
CALCULATION:
Given:
Length, l = 10 m
radius, r = \(\left(10^{-2} / \sqrt{\pi}\right) \)
and Electric resistance, R = 10 Ω
E = 10 V/m
According to the current density we have;
J = σE -----(1)
The resistivity is defined as the reciprocal of the conductivity, therefore, in equation (1) we have;
J = \(\frac{1}{ρ }\) E -----(2)
Now, resistivity, ρ =\(\frac{RA}{l}\)
Here, R is the resistance, A is the area of cross-section, and l is the length. Putting these values of resistivity in equation (2) we have;
J = \(\frac{l}{RA }\) E ----(3)
The area of cross-section ,A = \(\pi r^2\) = 10-4 m2
Now, on putting all the given values in equation (3) we have;
J = \(\frac{10}{10 \times 10^{-4} } \times 10\)
⇒J = 105 A/m2
Hence, option 1) is the correct answer.
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