Isn't the induced electric field vector always tangent to the loop?

Question

In the law of electromagnetic induction: $$\mathcal E_\text{ind}\equiv \oint_\Gamma \mathbf{E_\text{ }}d\mathbf l = -\dot \Phi_\mathbf{B}(t)$$ Here, $d\mathbf l = \hat{\mathbf t}dl$ is taken along the contour $\Gamma$. But isn't the field $\mathbf E$ which is responsible for the induced emf also along the contour (unit tangent vector $\hat{\mathbf t}$)?

Edit

Alright. From the comments and answers I have got I want to simplify something: the RHS can be decomposed to partial derivatives as, $$-\frac{\partial \Phi_\mathbf{B}(t)}{\partial t} +\oint_\Gamma \mathbf{(v×B)_\text{ }}d\mathbf l$$ Although the second term is due to Lorentz field which is not tangential, the first term indeed consists a tangential electric field: $$\oint_\Gamma E_\text{t}dl = -\frac{\partial \Phi_\mathbf{B}(t)}{\partial t} \leftrightarrow \text{curl }(E_\text{t}\hat{\mathbf t})=-\frac{\partial \mathbf B}{\partial t}$$ Regardless what orientation our loop is.

Answer 1

But isn't the field $\mathbf E$ which is responsible for the induced emf also along the contour (unit tangent vector $\hat{\mathbf t}$)?

No, the equation is true for any contour you choose in a given system. It may be convenient to choose a contour that does have the E field tangent at all points, but the equation is also true for other contours that aren't so chosen.

$$\mathcal E_\text{ind}\equiv \oint_\Gamma \mathbf{E_\text{ }}d\mathbf l = -\dot \Phi_\mathbf{B}(t)$$

You missed one key dot in the equation:

$$\mathcal E_\text{ind}\equiv \oint_\Gamma \mathbf{E_\text{ }}\cdot d\mathbf{l}$$

$\mathbf{E_\text{ }}\cdot d\mathbf{l}$ takes the magnitude of the component of $\mathbf{E}$ tangent to the contour element $d\mathbf{l}$; it doesn't require the field to be tangent to the contour.

Answer 2

Do you mean "which is responsible" or "that is responsible"? (See Strunk and White, Which Hunts: https://clubtroppo.com.au/2008/11/30/the-great-which-hunt/).

The former is descriptive, and since any electric field can have any vacuum solution E field added to it: the answer is: The direction of the electric field can't be specified, as there is no "the" electric field.

"That" is definitive, so it could specify just the field we're talking about, so let's assume it is.

You can imagine a nice cylindrically symmetric set up where $E$ always tangent to the $\Gamma$, and then put a kink in the wire. Field is not following that$^1$

Note also that the change in $\Phi$ isn't creating the $E_{ind}$, rather the charges and currents (and their derivatives) are creating both of them such that the curl of $E$ is proportional to the time derivative of $B$.

[1] ProfRob, PSE Comments, Jul 6, 2024 (Isn't the induced electric field vector always tangent to the loop?)