Spread the love

Current Density

The concept of current density is introduced to specify the current with direction at microscopic level at a point. “Current density is defined as a vector whose magnitude is equal to the magnitude of the current flowing per unit area normal to that point.“

Always remember that area is normal to the direction of charge flow (or current passes) through that point.

Current density at point is given by $\displaystyle \vec{J}=\frac{dI}{dA}\hat{n}$

If the cross–sectional area is not normal to the current, but makes an angle θ with the direction of current then current density $\displaystyle J=\frac{dI}{dA\cos \theta }$

$\displaystyle \Rightarrow dI=JdA\cos \theta$

$\displaystyle \Rightarrow dI=\overrightarrow{J}.d\overrightarrow{A}$

$\displaystyle \Rightarrow I=\int{\overrightarrow{J}.d\overrightarrow{A}}$

Current density $\displaystyle \vec{J}$ is a vector quantity. It’s direction is same as that of Electric Field Intensity $\displaystyle \vec{E}$ .

S.I. unit :- Ampere/m² & Dimension:- [L^–2A]

Note :-

If a steady current flows in a metallic conductor of non uniform cross section, then along the wire current (I) is same but current density (J) is different.

Here I₁ = I₂ , A₁ < A₂, J₁ > J₂

Example 1.
The current density at a point is $\displaystyle \vec{J}=\left( 2\times {{10}^{4}}\hat{j} \right)A{{m}^{-2}}$ . Find the rate of charge flow through a cross sectional area $\displaystyle \vec{S}=\left( 2\hat{i}+3\hat{j} \right)c{{m}^{2}}$ .

Solution :
The rate of flow of charge = current $\displaystyle I=\int{\overrightarrow{J}}.d\overrightarrow{S}=(2\times {{10}^{4}})\left[ \hat{j}.(2\hat{i}+3\hat{j}) \right]\times {{10}^{-4}}A=6A$

Example 2.
A potential difference applied to the ends of a wire made up of an alloy drives a current through it. The current density varies as J = 3 + 2r, where r is the distance of the point from the axis. If R be the radius of the wire, then find the total current through any cross section of the wire.

Solution :
Consider a circular strip of radius r and thickness dr

Current passing through strip of thickness dr, $\displaystyle dI=\overrightarrow{J}.d\overrightarrow{S}=(3+2r)(2\pi rdr)cos{{0}^{o}}=2\pi (3r+2{{r}^{2}})dr$

Total current passing through any cross section of the wire,

$\displaystyle I=\int\limits_{0}^{R}{2\pi (3r+2{{r}^{2}})dr=}2\pi \left( \frac{3{{r}^{2}}}{2}+\frac{2{{r}^{3}}}{3} \right)units$

Example 3.

Figure shows a conductor of length L carrying current I and having a circular cross – section. The radius of cross section varies linearly from a to b. Assuming that (b – a) << L. Calculate current density at distance x from left end.