A cylindrical region of radius R is filled with a uniform magnetic field B as shown in the figure. A metal wire (AB) of length L is placed inside the field such that its ends are symmetrically located with respect to the centre (O) of the circular cross section of the region. If the magnetic
field is changed at a rate $\frac{dB}{dt}$ the emf induced in the metal wire is $ε$ . Find change in value of e if the wire is displaced by a small distance $∆$ L parallel to its own length. Assume that the wire remains inside the field region.

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$\frac{△ L}{2} \frac{d B}{d t} L$

$\frac{△ L}{8 L} \frac{d B}{d t}$

$\frac{△ L}{4} \frac{d B}{d t} L$

$\frac{△ L}{2 L} \frac{d B}{d t}$

answer is A.

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Detailed Solution

The electric field induced at a distance r from the centre is given by
$\begin{array}{l} ∮ \vec{E} \vec{dl} = - \frac{dϕ}{dt} \\ \Rightarrow E 2 πr = {πr}^{2} \frac{dB}{dt} \Rightarrow E = \frac{r}{2} \frac{dB}{dt} \end{array}$
Component of E along AB is
$E_{AB} = Ecos θ = \frac{1}{2} (d / \cos θ) \cos θ \frac{dB}{dt} = \frac{d}{2} (\frac{dB}{dt}) = a constant .$
The emf induced in wire AB is $\int_{A}^{B} E \cos θ . d l$ = $\frac{d}{2} \frac{d B}{d t} \times (l e n g t h o f A B)$ .If the wire is displaced by a small amount $△ L$ parallel to itself, $E_{A B} w i l l c h a n g e b y \frac{△ L}{2} \frac{d B}{d t} L .$