Doubt about Faraday's Law

Question

Faraday's law states that

the circulation of the electric field E around a closed loop is equal to the rate of change of the magnetic flux through the area enclosed by the loop

In integral form:

$$\varepsilon =\oint \vec E \cdot d \vec l = - \frac{d\phi}{dt}$$

$\varepsilon$ is defined as the electromotive force, that is the work received by a unit charge as it goes around the closed circuit once. The electric field is a function of both time and space, $\vec E\,(\vec s, t)$.

Given the above definition of e.m.f., as a unit charge travels around the closed circuit time passes and the electric field varies as a consequence of this, since it is a function of time.

As far as my understanding goes, instead the line integral $ \oint \vec E \cdot d\vec l $ treats the electric field as constant with respect to time all along the path. So, to me, it seems like this isn't the real work received by a unit charge travelling around the closed loop.

Shouldn't this time-dependency be taken into account when computing the line integral?

Answer 1

So, to me, it seems like this isn't the real work received by a unit charge travelling around the closed loop.

Yes, that's right - unless of course the emf is constant in time.

Shouldn't this time-dependency be taken into account when computing the line integral?

If you're trying to calculate the work done on an actual charge which is physically traveling around the loop, then yes. If you're making a statement about the relationship between the circulation of the electric field and the rate of change of the magnetic flux at a particular instant of time, then no.

Faraday's law is ultimately the statement that the circulation of the electric field at some point $\mathbf x$ and some time $t$ is equal to (minus) the rate of change of the magnetic field at $(\mathbf x,t)$. In differential form, this reads

$$\bigg(\nabla \times \mathbf E\bigg)(\mathbf r,t) = -\frac{\partial \mathbf B}{\partial t}(\mathbf r,t)$$

This is true at every point $\mathbf r$ and time $t$.

Alternatively, this expression can be integrated over some spatial surface $S$ which is bounded by a loop $\partial S$, with the result being

$$\oint_{\partial S} \mathbf E(\mathbf r,t) \cdot d\mathbf r = -\frac{d}{dt} \int_S\mathbf B(\mathbf r,t) \cdot d\mathbf S$$

Both the line integral and the time derivative of the surface integral are computed at fixed time $t$.

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The electrical work done on a point charge actually moving around the loop would be

$$\oint \mathbf E\bigg(\mathbf r(t),t\bigg) \cdot \frac{d\mathbf r}{dt} dt$$

Contrast this with how you'd evaluate the integral at fixed time:

$$\oint \mathbf E\bigg(\mathbf r(\lambda),t\bigg) \cdot \frac{d\mathbf r}{d\lambda} d\lambda$$

where we've parameterized our loop with parameter $\lambda$. In the former case, the time argument of the electric field changes, but in the latter case it remains fixed. If $\mathbf E$ is constant in time, then the two expressions of course coincide.

Answer 2

The "unit of charge" has to be very small, small enough not to affect the E field as it travels around the loop. Then the E field is time-invariant (for a constant-ramp dB/dt) =, and the work done by a differential charge dq is $ dq \oint \bf E \cdot \bf{dl} = dq \times emf $.

Even for randomly-time-varying B fields resulting in time-varying E fields, the work done by a small charge dq around the loop is still emf x dq.

Answer 3

actually when you are calculating line integral you assume electric field to be constant with respect to time for a small period oftime and using this the result is not affected much. so for a time varying electric field we assume that electric field is constant for the time period you are calculting integral