please i want to ask when induced current produced by changing magnetic field according to faraday's law is this current DC or AC current ?


1 Answer 1


Faraday's law of induction can be written as

$$\oint\limits_C \vec{E}\text{ }d\vec{l} = - \frac{d}{dt}\int\limits_A \vec{B}\text{ }d\vec{A}$$

where $C$ is some closed curve and $A$ is the area bounded by it. Recalling that the electric field is the gradient of the electric potential (i.e. something like a derivative) and that voltage is nothing but a potential difference, you will see that the integral of the electric field will first of all lead you to an induced voltage. Usually, due to Ohm's law, the voltage will be proportional to the current that you are looking for. The answer to your question is then simply that the voltage (and thereby the current) will change in the same manner in which $\frac{d}{dt}\vec{B}$ changes. Thus, if the magnetic flux density increases or decreases constantly, its time derivative will be constant and the induced voltage (current) will be constant (DC) as well. In praxis, however, one will often face sinusoidal functions, a simple example of which could be written as $$|\vec{B}|=B_0 \sin{\omega t}$$ As the time derivative of such a function is again a sinusoidal function (in this case a cosine), the resulting voltage and current will therefore also be alternating. (Of course, one can always complicate things with messy functions for $\vec{B}$, non-uniform distributions over $A$ and so on but the important part to note here from a didactic point of view is the dependence on the time derivative of $\vec{B}$ and not on $\vec{B}$ itself or whatever).


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