Question 44: Electromagnetic Induction

Question

Two metallic rings A and B, identical in shape and size but having different resistivities $$\rho_A$$ and $$\rho_B$$, are kept on top of two identical solenoids as shown in the figure below. 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)

Accepted Answer

Induced emf is same in both the rings:
$$I = {e \over R} = {{\rho A} \over {\rho l}}$$
$$I \propto {1 \over \rho } \Rightarrow q \propto {1 \over \rho }$$ .... (1)
Impulse is
$$J = \int {Bil\,dt = mv = Bl\int {I\,dl = mv} } $$
That is, $$J = Blq = mv \Rightarrow v\left( {{q \over m}} \right)$$ ..... (2)
and $${v^2} \propto h$$ ...... (3)
From Eqs. (1), (2) and (3), we get
$$mv \propto {1 \over \rho }$$
$$m\sqrt h  \propto {1 \over \rho }$$
$$m\rho  \propto {1 \over {\sqrt h }}$$
Since $${h_A} > {h_B}$$ and for $${m_A} = {m_B},\rho \sqrt h  = $$ Constant. Therefore, $${\rho _A}  {h_B}$$ (already given)

Answer

$$\rho_A$$ > $$\rho_B$$ and $$m_A$$ = $$m_B$$

Answer

$$\rho_A$$ < $$\rho_B$$ and $$m_A$$ = $$m_B$$

Answer

$$\rho_A$$ > $$\rho_B$$ and $$m_A$$ > $$m_B$$

Answer

$$\rho_A$$ < $$\rho_B$$ and $$m_A$$ < $$m_B$$

Electromagnetic Induction

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