Anarrow 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 field in the wide wire is

(a) e
(b) 2e
(c) 4e
(d) e/2
(e) e/4

E) E₂=E/4

Explanation:

Given that

Both wire are connects in the series so the current flow will be same in the both wire but voltage difference will be different.

Resistance R

1 - for first wire

2 - for second wire

R₁=ρL/A                

R₂=ρL/A'

Given that radius of second wire is double then the area of second wire will be 4 times of first wire.

A'= 4 A

R₂=ρL/A'

R₂=ρL/4A

R₂ = R₁/4

For wire 1

We know that

V= I R

V₁= I R₁

V₂=I R₂

R₂ = R₁/4

so

V₁=4 V₂

So we can say that

E₂=E₁/4

Given that E₁=E

E₂=E/4

(75°f − 32) × 5/9 = 23.889°c