The relationship between initial and final current density

[Fatima Hasan]

Homework Statement

Homework Equations

$J=\frac{I}{A}$
A=π r²

The Attempt at a Solution

Since I is constant and the diameter is doubled , A₂ = π ($\frac{2D}{2}$)²
A₂ = 4 A₁
J₂ = I/(4A₁)
= J₁ /4 → (E)
Is my answer correct ?

 

[cnh1995]

Since I is constant

Are you sure?

 

[Fatima Hasan]

Are you sure?

$R=\frac{ρL}{A}$
The area goes up by 4 if the diameter is doubled. So, R₂ = R₁ / 4
$I=\frac{ΔV}{R}$ ( I is inversely proportional to R )
I goes up by 4 . (I₂ = 4I₁)
J = I / A
J₂ = $\frac{4I}{4A}$
J₂ = J₁

 

[cnh1995]

$R=\frac{ρL}{A}$
The area goes up by 4 if the diameter is doubled. So, R₂ = R₁ / 4
$I=\frac{ΔV}{R}$ ( I is inversely proportional to R )
I goes up by 4 . (I₂ = 4I₁)
J = I / A
J₂ = $\frac{4I}{4A}$
J₂ = J₁

Right.

 

[rude man]

The problem can't be solved since the diameters (or ratio of diameters) is not given.
(I realize the assumption of D₂ = 2D₁ is made here).

 

[haruspex]

The problem can't be solved since the diameters (or ratio of diameters) is not given.

No, it is irrelevant, though all the wrong answers seem to assume the ratio is 2:1.