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(B) When the current $I$ is switched on in the solenoid,the magnetic flux through the ring changes,inducing an electromotive force $(EMF)$ $\varepsilo

Vedclass · Accepted Answer

$(B, D)$

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(B) When the current $I$ is switched on in the solenoid,the magnetic flux through the ring changes,inducing an electromotive force $(EMF)$ $\varepsilon = -\frac{d\phi}{dt}$.The induced current in the ring is $i = \frac{\varepsilon}{R} = \frac{1}{R} \frac{d\phi}{dt}$,where $R = \rho \frac{L}{A_{cs}}$ is the resistance of the ring ($L$ is the circumference,$A_{cs}$ is the cross-sectional area of the wire).The magnetic force on the ring is $F = i L B_r$,where $B_r$ is the radial component of the magnetic field. Since $i \propto \frac{1}{\rho}$,the impulse $J = \int F dt = \int i L B_r dt \propto \frac{1}{\rho} \int \frac{d\phi}{dt} dt = \frac{\Delta \phi}{\rho}$.By the impulse-momentum theorem,$J = m v$,so $v = \frac{J}{m} \propto \frac{1}{\rho m}$.The height reached is $h = \frac{v^2}{2g} \propto \frac{1}{\rho^2 m^2}$.Given $h_A > h_B$,we have $\frac{1}{\rho_A^2 m_A^2} > \frac{1}{\rho_B^2 m_B^2}$,which implies $\rho_A m_A Checking the options:$(A)$ If $\rho_A > \rho_B$ and $m_A = m_B$,then $\rho_A m_A > \rho_B m_B$ (Incorrect).$(B)$ If $\rho_A $(C)$ If $\rho_A > \rho_B$ and $m_A > m_B$,then $\rho_A m_A > \rho_B m_B$ (Incorrect).$(D)$ If $\rho_A Thus,the possible relations are $(B)$ and $(D)$.

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