A 'twisted' circular band (called a Moebius strip) is made from a strip of paper of length $L$ and width $d$. A wire running along the edge of the strip is connected to a voltmeter, as shown in the figure.

What does the voltmeter register when the strip is placed in a homogeneous magnetic field which is perpendicular to the plane of the strip and changes uniformly with time, i.e. $B\left(t\right)= kt$?

Consider two touching discs each of radius $R=L/2\pi$ as shown in Fig.(a).

If the discs are placed with their plane perpendicular to a homogeneous magnetic field whose strength changes uniformly with time, the voltage induced in the piece of wire wrapped around their edge is 
$V=2\pi R^2\frac{\Delta B}{\Delta t}.$

Now twist the disc on the right by $180°$ about its symmetry axis e (Fig (b)(i)). Its top (dark) side then becomes its bottom side (Fig (b)(ii)). Turn the same disc again by $180°$, but this time about axis $t$ (Fig (b)(iii)). At the end of this, the dark side of both discs are on top and the perimeter is exactly the same as that of the Moebius strip mentioned in the problem. 
Thus, with the strip in a uniformly changing magnetic field, the voltmeter reads 
$V=2\pi R^2\frac{\Delta B}{\Delta t}=\frac{k{L}^2}{2\pi }$
This value is much higher than what one would natively expect, if reasoning from the area of the paper band. The area of the (one-sided) surface covering the Moebius strip is not the same as the area of the paper band, and for narrow strips it is, in fact, much larger! The induced voltage can also be calculated by cutting the wire at the ‘twist’ into two ‘coils’ of one turn each, and adding up the algebraic values of the voltages induced in each turn (taking account of their directions). In the present case, the directions of the two turns are the same, and therefore the voltage $V_0=k\pi R^2$ in one turn is doubled to give a total voltage of $V=2V_0$.