A rectangular loop of wire is situated so that one end (height h) is between the plates of a parallel-plate capacitor (Fig. 7.9), oriented parallel to the field E. The other end is way outside, where the field is essentially zero. What is the emf in this loop? If the total resistance is R, what current flows? Explain. [Warning: This is a trick question, so be careful; if you have invented a perpetual motion machine, there's probably something wrong with it.]

The electric field is E.

Th total resistance is R.

The height between the plates is h.

The emf of the loop is given by,

$\mathbf{\epsilon }\mathbf{=}\mathbf{\int }\mathbf{E}\mathbf{-}\mathbf{dI}$ …… (1)

For all the electrostatic field, the emf is equal to zero.

Therefore,

$\mathbf{\int }\mathbf{E}\mathbf{.}\mathbf{dI}\mathbf{=}\mathbf{0}$

The electric field due to the length of the loop parallel to the capacitor plate is zero because the $\stackrel{\rightharpoonup }{dI}$is perpendicular to the electric field. The electric field outside the capacitor is zero. Therefore, the emf due to the wire containing the resistance is zero. The electric field is existed only because of the wire inside the capacitor and parallel to the plates.

The field inside the conductor is $E=\frac{\sigma }{{\epsilon }_0}$

Substitute the $\frac{\sigma }{{\epsilon }_0}$for and for into equation (1).

$\epsilon =\frac{\sigma }{{\epsilon }_0}h$

The electric field is not constant inside the capacitor, and due to fringing, not all the field lines are perpendicular to the field; therefore, the electrostatic field inside the capacitor becomes zero. Therefore, the emf of the loop is zero.

$\epsilon =0V$

The current in the loop is given by,

$I=\frac{\epsilon }{R}$

Substitute 0V for $\epsilon$into above equation.

$$\begin{gathered} I=\frac{0}{R} \\ I=0A \end{gathered}$$

Hence, the emf inside the current is 0V and current is 0A.