A highly conducting iron cylindrical annulus with permeability $μ$ and inner and outer radii $R_{i}$ (shown as $R_{1}$ in fig) and $R_{o}$ (shown as $R_{2}$ in fig) respectively, is placed concentrically to an infinitely long straight wire carrying a direct current I as shown in figure. The magnetic flux density is B at a point and distances are measured from the axis of cylinder.
Case-1: A highly conducting circuit abcd is moving down wards with a constant velocity $v_{0}$ , while making contact with the surfaces of the cylindrical annulus through sliding brushes. The circuit is completed from c to d through the iron cylinder. The induced emf in this case is $ξ_{1}$ .
Case-2: Following from the information of case-1, now the circuit remains stationary, while the cylinder moves upwards with the same velocity $v_{0}$ . The induced emf in this case is $ξ_{2}$ .
Case-3: A thin axial slot is cut in the cylinder, so that the circuit abcd can be formed completely by wire and can slide in the slot. The circuit is kept fixed and the cut cylinder moves up wards with constant velocity $v_{0}$ . The induced emf in this case is $ξ_{3}$ .
Then

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$B = \frac{μ_{0} I}{2 π r}$ for $r < R_{i}$ and $B = \frac{μ I}{2 π r}$ for $R_{i} < r < R_{o},$ in the tubing

$| ξ_{1} | = \frac{μ_{0} I V_{o}}{2 π} \ln \frac{R_{o}}{R_{i}} = | ξ_{2} |$

In Case-1 and Case-2 emf induced is independent of $μ$

$| ξ_{3} | = \frac{(μ - μ_{0}) I V_{o}}{2 π} \ln \frac{R_{o}}{R_{i}}$

answer is A, B, C, D.

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Detailed Solution

$d ϕ = \frac{μ_{0} I}{2 π r} y d r ϕ = \frac{μ_{0} I}{2 π} y \ln (\frac{R_{2}}{R_{1}}) \frac{d ϕ}{d t} = \frac{μ_{0} I v_{0}}{2 π} \ln (\frac{R_{2}}{R_{1}}) = \in_{1} = \in_{2} ϕ_{1} = \int \frac{μ I}{2 π r} d r (l - y) ϕ_{2} = \frac{μ_{0} I}{2 π} \ln (\frac{R_{2}}{R_{1}}) y ϕ_{1} = \frac{μ I}{2 π} (l - y) \ln (\frac{R_{2}}{R_{1}}) ϕ = \frac{μ I}{2 π} \ln (\frac{R_{2}}{R_{1}}) (l - y) + \frac{μ_{0} I}{2 π} \ln (\frac{R_{2}}{R_{1}}) y ξ_{3} = \frac{d ϕ}{d t} = \frac{I \ln (\frac{R_{2}}{R_{1}})}{2 π} [μ_{0} - μ] v_{0}$