# Difference in induced current, when magnetic field “span” is reduced?

A conductor of known volume $(V)$ passes a uniform magnetic field$(B)$with a constant velocity $(v)$ the conductor is a source of induced EMF, a power source to a circuit. The induced EMF can be calculated via formula: $$\epsilon = -vBL$$

A diagram of this:

The conductor has resitance$(R)$ and induces current equal to:

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

Excluding all other factors that would retard the motion, what would happen if the magnetic field span is reduced? Span; meaning It covers a smaller volume(i.e area) of the conductor at the same magnetic field strength as before, moving at the same velocity:

From the motional EMF formula, if all the vairbales are the same $(v,B,L)$ it should induce the same $\epsilon$, however, the induced current would be less? Although the resistance is still the same...? Or does one of the formulas fail/incorrect in my assumptions?

Here is where I think, ohm's law in calculating the induced current would be invalid?

The voltage V that you're computing would be induced between the upper and lower sides of your orange block, and $L$ in your computation is the distance between these two sides. This voltage is induced between any two opposite points on the upper and lower sides provided they are in the area that is covered by the magnetic field. If you connect a wire with resistance $R$ between these points, then you'll get current $I=V/R$ out. Thus, what changes if you decrease the area covered by the magnetic field, is that the points where the wire is connected will leave the field quicker, and you will draw current during a smaller amount of time.
If you use the whole upper and lower sides as thick wires, then the relevant resistance is $R=\sigma/l$ where $\sigma$ is an appropriate resistivity and $l$ the span of the block that is covered by the magnetic field. Then the current is given by $I=V/R=V*l/\sigma$. From here you see that the smaller the span $l$ the less current will be drawn.
• More focused about using the whole upper and lower side, focusing on $V$ prior to $I$, it would be less than the first figure? $V$ = $V_i$ * $Z_2$/ $(Z_1 + Z_2)$? Where Z2 is the resistance outside the magnetic field, Z1 is the resistance inside the magnetic field. Voltage is divide due to the parts of the conductor outside the magnetic field? – Pupil Aug 8 '15 at 4:49
• As you said yourself, as long as the parameters $(v, B, L)$ are fixed, the induced voltage $V$ is also fixed in terms of these parameters. – Stan Aug 8 '15 at 5:59
• yep:) as long as $B$ is not equal zero. – Stan Aug 11 '15 at 9:21