Question

f.

The figure below represents a section of a circular conductor of nonuniform diameter carrying a current of I = 5.90 A. The radius of cross-section A, is r 1 = 0.580 cm. 12 (a) What is the magnitude of the current density across A7? | A/m² The radius raat Az is larger than the radius rı at A1 (b) is the current at A larger, smaller, or the same? The current is larger. O The current is smaller. O The current is the same. (c) Is the current density at A, larger, smaller, or the same? O The current density is larger. O The current density is smaller. O The current density is the same. Assume Az = 341 (d) Specify the radius at Az: mm (e) Specify the current at Az.

f Specify the current density at A₂.

Answer 1

a)

We have the current passing through area of cross section $A_1$ is,

11 = 5.90A

The radius of this cross section $A_1$ is,$r_1=0.580cm=0.58*10^{-2}m$

So,The area of cross section of $A_1$ is,$A_1=\pi{r_1}^2=\pi(0.58*10^{-2})^2=1.057*10^{-4}m^2$

So,The current density across $A_1$ is,$J_1=\frac{I}{A_1}=\frac{5.90}{1.057*10^{-4}}=55818.353A/m^2$

$J_1=55818.353A/m^2$

b)

Here even though there is a change in area of cross section for the conductor,since the current is the flow of an amount of electron per unit time will be same through the cross section $A_2$ as that of $A_1$ .

So,The same current coming from $A_1$ will passes through $A_2$ ,or the current through the conductor here remains the same.

So,The current is same

c)

We have the current density is the ratio of current to cross sectional area,ie,$J=\frac{I}{A}$

Here the current through $A_1$ and $A_2$ are same or constant.

So,current density is inversely proportional to the area of cross section. ie,$J\alpha \frac{1}{A}$

Since the Area of cross section $A_2$ is larger than the area of cross section $A_1$ .

So,The current density through $A_2$ is smaller than the same through $A_1$ .

ie,the current density is smaller.

d)

The radius of this cross section $A_1$ is,$r_1=0.580cm=0.58*10^{-2}m$

So,The area of cross section of $A_1$ is,$A_1=\pi{r_1}^2=\pi(0.58*10^{-2})^2=1.057*10^{-4}m^2$

Also given that,$A_2=3A_1$

So,The area of cross section,$A_2=3A_1=3*1.057*10^{-4}=3.171*10^{-4}m^2$

ie,$A_2=3.171*10^{-4}m^2$

Let the radius of $A_2$ be $r_2$

So,$A_2=\pi{r_2}^2$

So,$r_2=\sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{3.171*10^{-4}}{\pi}}=1.0046*10^{-2}m=10.046mm$

Or radius of cross section $A_2$ , $r_2=10.046mm$

e)

Since the current is same for $A_1$ and $A_2$ equals to,$I=5.90A$

So,The current at $A_2$ is,$I=5.90A$

f)

We have the area of cross section at $A_2$ is,$A_2=3A_1=3*1.057*10^{-4}=3.171*10^{-4}m^2$

Or,$A_2=3.171*10^{-4}m^2$

Also the current through $A_2$ is,$I_2=I_1=I=5.90A$

So,the current density through $A_2$ ,$J_2=\frac{I}{A_2}=\frac{5.90}{3.171*10^{-4}}=18606.118A/m^2$

The current density through $A_2$ ,  $J_2=18606.118A/m^2$

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