A ring of mean radius $r$ is made of a perfectly conducting wire of cross-sectional area $A$ . Inductance $L$ of the ring is so small and inertia of free electrons cannot be neglected in the current-building process. The free electron density in the conductor is $n$ , mass of an electron is $m$ and modulus of charge on an electron is $e$ . Initially the ring is placed in a uniform magnetic field with its plane parallel to induction vector $\bar{B}$ of the field as shown in figure. Find the current in ring after it is turned through an angle $90^{0}$ .

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$i = \frac{B \times 2 π r}{L + \frac{m \times π r^{2}}{n e^{2} A}}$

$i = \frac{B π r^{2}}{L + \frac{m \times π r}{n e^{2} A}}$

$i = \frac{B π r}{L + \frac{m \times π r}{n e^{2} A}}$

$i = \frac{B π r^{2}}{L + \frac{m \times 2 π r}{n e^{2} A}}$

answer is C.

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Detailed Solution

For perfect conductor, R=0

Initially, $i = 0 & \vec{B} ⊥ \vec{A}, ϕ_{i} = B A \cos 90 ° = 0$
As the ring rotates, induced emf is developed in it due to change in magnetic flux through it.
Induced current, $i = \frac{B π r^{2}}{L_{net}} = \frac{B π r^{2}}{L + \frac{m \times 2 π r}{n e^{2} A}}$ (using $∆ ϕ = L ∆ i$ )
Finally, $ϕ_{f} = - B A + L_{net} i$