Current has both magnitude and direction. As per the definition of vector defined in encyclopedia, current should be a vector quantity. But, we know that current is a scalar quantity. What is the reason behind it?
[Physics] Why is current a scalar quantity
electric-currentelectromagnetismvectors
Related Solutions
According to my understanding, indeed you could define a physical quantity like $$\vec{I} = I\;\vec{n_d}$$ where $\vec{n_d}$ is the unitary drift direction. There is no problem with that. But what is the most important is to understand the harmony between different quantities. I mean that there is some little subtleties between $I$ and $\vec{j}$.
The current is defined according to a surface $A_t$ (and is a local quantity: the position of the surface). Since the surface may be tilted (not perpendicular to $\vec{n_d}$), then in general, we should write $$I=n\;q \; \vec{A_t}.\vec{v_d}=n\;q\;A_t \; \vec{n_{A_t}}.\vec{v_d}=n\;q\;A_t \; cos(\theta).v_d$$ where $\theta$ is the angle between $\vec{A_t}$ and $\vec{v_d}$ (If the charge $q$ is negative, $\vec{I}$ and $\vec{n_d}$ would have opposite orientations, which fits with the usual convention of the electric current). Of course, if we have considered a surface $A$ which is perpendicular to the drift, the current would be the same but we would write $$I=n\;q \; \vec{A}.\vec{v_d}=n\;q\;A \; v_d$$
That is, let's talk about the densities. We have $$\mathrm{d}I=n\;q\;v_d\;\mathrm{d}A=n\;q\;v_d\;cos(\theta)\;\mathrm{d}A_t$$ The scalar current density is given by $$j=\frac{\mathrm{d}I}{\mathrm{d}A}=n\;q\;v_d=\frac{1}{cos(\theta)}\frac{\mathrm{d}I}{\mathrm{d}A_t}$$ Yet, you see that you have to be careful about the differential area that you put in the denominator. The current could then be calculated as $$I=\int_A j\;\mathrm{d}A=\int_{A_t} j\;cos(\theta)\;\mathrm{d}A_t$$ Here also, we see a possible source of confusion. This can be fixed if we define a vector current density as $$\vec{j}=j\;\vec{n_d}=n\;q\;v_d\;\vec{n_d}$$ The direction of $\vec{j}$ is resolved too: it's that of the local drift in the considered material position. That direction may be different than that of the average drift of the whole current $\vec{I}$. That is, the expression of the current can be written simply as $$I=\int_A \vec{j}\;\mathrm{d}\vec{A}=\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}$$ Or, you could use your own convention and write $$\vec{I}=\bigg(\int_A \vec{j}\;\mathrm{d}\vec{A}\bigg)\;\vec{n_d}=\bigg(\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}\bigg)\;\vec{n_d}$$
That definition of a vector quantity is a little too simple. It needs to not only have direction, but the directions need to add depending on the angles between them in a specific way to give an overall equivalent quantity.
Current in a circuit isn't really a vector quantity, it has direction but that is equivalent to just the sign of the current. You can have a positive current going in one direction and a negative current in the other - they will still add but not in a vector sense.
Perhaps there needs to be a 3rd term in between scalar and vector.
Best Answer
To be precise, current is not a vector quantity. Although current has a specific direction and magnitude, it does not obey the law of vector addition. Let me show you.
Take a look at the above picture. According to Kirchhoff's current law, the sum of the currents entering the junction should be equal to sum of the currents leaving the junction (no charge accumulation and discharges). So, a current of 10 A leaves the junction.
Now take a look at the picture below.
Here, I have considered current to be a vector quantity. The resultant current is less than that obtained in the previous situation. This result gives us a few implications and I would like to go through some of them. This could take place due to charge accumulation at some parts of the conductor. This could also take place due to charge leakage. In our daily routine, we use materials that are approximately ideal and so these phenomena can be neglected. In this case, the difference in the situations is distinguishable and we cannot neglect it.
If you are not convinced, let me tell you more. In the above description (current as a vector), I have talked about the difference in magnitudes alone. The direction of the resultant current (as shown) is subtle. That's because in practical reality, we do not observe the current flowing along the direction shown above. You may argue that in the presence of the conductor, the electrons are restricted to move along the inside and hence it follows the available path. You may also argue that the electric field inside the conductor will impose a few restrictions. I appreciate the try but what if I remove the conductors? And I also incorporate particle accelerators that say shoot out proton beams thereby, neglecting the presence of an electric field in space.
Let me now consider two proton beams (currents), each carrying a current of 5 A as shown below. These beams are isolated and we shall not include any external influences.
Now that there is no restriction to the flow of protons, the protons meeting at the junction will exchange momentum and this will result in scattering (protons represented by small circles). You would have a situation where two beams give rise to several beams as shown below. Our vector addition law does not say this.
I have represented a few in the picture above. In reality, one will observe a chaotic motion. Representation of the beams (as shown right above) will become a very difficult task because the protons do not follow a fixed path. I have just shown you an unlikely, but a possible situation.
All this clearly tells us that current is not a vector quantity.
Another point I would like to mention is, current cannot be resolved into components unlike other vector quantities. Current flowing in a particular direction will always have an effect along the direction of flow alone over an infinite period of time (excluding external influences such as electric or magnetic fields).