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A ring of radius a = 50 mm made of thin wire of radius b = 1.0 mm was located in a uniform magnetic field with induction B = 0.5 mT so that the ring plane was perpendicular to the vector B. Then the ring was cooled down to a super-conducting state, and the magnetic field was switched off. Find the ring current after that. Note that the inductance of a thin ring along which the surface current flows is equal to $L = μ_{0} a (\ln (\frac{8a}{b}) - 2)$

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20 A

50 A

40 A

15.658A

answer is D.

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Detailed Solution

induced emf

e = \frac{d ϕ}{dt} \Rightarrow L \frac{d i}{dt} = \frac{d ϕ}{dt}

$\Rightarrow L i = ϕ$

$\Rightarrow i = \frac{π r^{2} B}{L}$

$i = \frac{π a^{2} B}{μ_{0} a [l n (\frac{8a}{b} - 2)]}$

$= \frac{π aB}{4 π \times 10^{- 7} [l n (\frac{8 \times 50 \times 10^{- 3}}{10^{- 3}}) - 2]}$

$= \frac{50 \times 10^{- 3} \times 5 \times 10^{- 4}}{4 \times 10^{- 7} [l n 400 - 2]} = 50 A$

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