In: Physics
A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narrow wire is E, the electric firld in the wide wire is.
As we know that Current in both wire will be same but both have different resistance as per
\(\mathrm{R} 1=\rho_{\mathrm{L} / \mathrm{A}}\)
\(\mathrm{R} 2=\rho_{\mathrm{L} / 4 \mathrm{~A}}\) (Because redius is double so cosssectional area will be four times)
So \(\mathrm{R} 1=4^{*}(\mathrm{R} 2)\)
So if current is same in both then voltage will be -
\(\mathrm{V} 1=4^{*}(\mathrm{~V} 2)\) (because \(\vee\) propostional \(\mathrm{R}, \mathrm{i}\) is same in all)
then-
\(E 1=4(E 2)\) (because E propostional \(V, r\) is same in all)
E2 \(=(E 1) / 4\) (Electric feild in wide wire)
Option E.