Does the electric current have a direction?

[hmalkan]

We know that the electrical current is scalar. Also we know that a scalar hasn't got a direction but electric current has got a direction. I've confused! Please help me..

 

[ZapperZ]

We know that the electrical current is scalar. Also we know that a scalar hasn't got a direction but electric current has got a direction. I've confused! Please help me..

How do you know that electric current is a scalar?

Zz.

 

[csprof2000]

Electric current does have a direction, and, as such, is a vector, and not a scalar.

 

[confinement]

Electic current does have a direction, the problem is the frequent appearance of the equation:

$$i = \frac{dq}{dt}$$

which does not really tell exactly what current is; this equation only gives you the magnitude, not the direction. I prefer to define the surface current density:

$$\vec{j} = \rho \vec{v}$$

in terms of the charge density and the velocity. Then we have:

$$\vec{i} = A \vec{j}$$

where A is the area.

 

[hmalkan]

http://ecx.images-amazon.com/images/I/41VHYYJB0KL._SL160_.jpg

This is my book. It says electric current is a scalar at the heading of chapter 2.

​

 

[jtbell]

The current density $\vec J$ is a vector, but the current I through a given surface is a scalar, as can be seen from the relationship between the two:

$$I = \int{\vec J \cdot d \vec a}$$

When you're calculating e.g. the magnetic force on a current-carrying wire, the directionality of the current is properly associated with the length of the wire rather than with the current itself:

$$\vec F = I \vec l \times \vec B$$

for a straight wire segment and uniform $\vec B$, or

$$\vec F = I \int {d \vec l \times \vec B}$$

otherwise. This assumes that $\vec B$ doesn't vary significantly over the cross-section of the wire. If it does, then you have to calculate the force by using the current density and integrating over the volume of the wire:

$$\vec F = \int {(\vec J \times \vec B) dV}$$

 

[jtbell]

$$\vec{i} = A \vec{j}$$

where A is the area.

You have to allow for the area not being perpendicular to the current. If $\vec J$ is uniform, then you can use

$$I = \vec J \cdot \vec A$$

where the direction of $\vec A$ is perpendicular to the surface. If $\vec J$ is not uniform, then you have to integrate.

 

[hmalkan]

The current density $\vec J$ is a vector, but the current I through a given surface is a scalar, as can be seen from the relationship between the two:

$$I = \int{\vec J \cdot d \vec a}$$

When you're calculating e.g. the magnetic force on a current-carrying wire, the directionality of the current is properly associated with the length of the wire rather than with the current itself:

$$\vec F = I \vec l \times \vec B$$

for a straight wire segment and uniform $\vec B$, or

$$\vec F = I \int {d \vec l \times \vec B}$$

otherwise. This assumes that $\vec B$ doesn't vary significantly over the cross-section of the wire. If it does, then you have to calculate the force by using the current density and integrating over the volume of the wire:

$$\vec F = \int {(\vec J \times \vec B) dV}$$

I don't understand what $\vec B$ stands for.

 

[Vanadium 50]

I don't understand what $\vec B$ stands for.

Magnetic field.

 

[Archosaur]

http://ecx.images-amazon.com/images/I/41VHYYJB0KL._SL160_.jpg

This is my book. It says electric current is a scalar at the heading of chapter 2.

​

Current is scalar.

Current (in amperes) is the amount of charge that passes through a point on a conductor every second. It is just a number. An "ampere" is a scalar quantity.

But, and this might make it confusing, but the electrons do flow in a given direction.
The electrical current does have a direction, but that information isn't contained in the unit "ampere".

Does that make sense at all? I could try to explain it better...

 

[csprof2000]

The units only ever measure magnitude, not direction.

It doesn't even make sense to have vector units.

 

[hmalkan]

All you are very helpful. Thanks for replies.