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Figure(a): enter image description here

When a conductor moving inside magnetic field, of a given length at a certain velocity the induced EMF is:

$$\epsilon = vBL$$

However, what if we changed the position where the bottom wire is connected to the wire like so: enter image description here

Is the induced EMF now:

$$\epsilon = vBL_2$$

I'm not sure how that can be true, when the conductor's length has not changed just the position of where the circuit wire is connected "shortining" the current medium(or path) I agree, however, how would it change the induced EMF? The charges are still at the top & bottom of the conductor, would the terminal wire's connection reduce the induced EMF?

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The magnetic flux intercepted by the conductor will be reduced by connecting the wire at intermediate point of conductor because the effective length of the conductor will now $L_2$ & induced E.M.F. is given as $$\epsilon =vBL_2$$

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  • $\begingroup$ That is my initial thoughts as well, however, the effected length for induced EMF should be $L$ it's the medium moving inside the magnetic field, the top part(having all the positive charge is connected) to the circuit. $\endgroup$ – Pupil Dec 4 '15 at 5:49
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You need to look at the rate of change of flux through the circuit, including the flux being added by the slanted section of wire (stretchable?). A more realistic (and simpler) arrangement would be to have a rod sliding on parallel rails which can be moved closer together.

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We know that

$\mathcal{E}=\oint \mathbf{f}_{\operatorname{mag}} \cdot d \mathbf{l}$ with $f_{\text {mag }}$ being magnetic force per unit charge. where the integral is taken around the loop through which the current flows.

Since outside our conducting Bar the integral reduceds to zero we have then

$\mathcal{E}=\int_{0}^{L_{2}} f_{\operatorname{mag}} \cdot d l$

$\Rightarrow$ $\mathcal{E}= v B L_2$

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