JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Electromagnetic Induction

Two metallic rings A and B, identical in shape and size but having different resistivities

ρ_{A}

and

ρ_{B}

, are kept on top of two identical solenoids as shown in the figure. When current

I

is switched on in both the solenoids in identical manner, the rings A and B jump to heights

h_{A}

and

h_{B}

, respectively, with

h_{A} > h_{B}

. The possible relation(s) between their resistivities and their masses

m_{A}

and

m_{B}

is(are)

(A) $ρ_{A} > ρ_{B}$ and $m_{A} = m_{B}$
(B) $ρ_{A} < ρ_{B}$ and $m_{A} = m_{B}$ ✓ Correct
(C) $ρ_{A} > ρ_{B}$ and $m_{A} > m_{B}$
(D) $ρ_{A} < ρ_{B}$ and $m_{A} < m_{B}$ ✓ Correct

Answer: (B), (D)

Solution

When current is switched on in the solenoids, the changing magnetic flux induces an emf in the rings. Since the emf is the same for both rings (identical solenoids), the induced current depends on the resistance of the ring. Lower resistivity means lower resistance and higher induced current. Higher induced current leads to greater repulsive force and thus greater height. Since

h_{A} > h_{B}

, ring A must have lower resistivity:

ρ_{A} < ρ_{B}

. This can occur with equal masses (option B) or if ring A is lighter (option D).

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · One or more correct options

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JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Electromagnetic Induction

Solution

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