Levitation of Conducting Ring Above AC Powered Coil

To determine the correctness of the statements, let us analyze the electromagnetic interaction between the AC-powered coil, the iron rod, and the conducting ring (the classic Thomson's jumping ring experiment):

1. Magnetic Field Distribution:
   When an alternating current $I(t) = I_0 \cos(\omega t)$ passes through the coil, it produces an alternating magnetic field channeled through the iron rod. Above the coil, the magnetic field lines diverge as they exit the rod. In cylindrical coordinates $(r, \theta, z)$, this divergence implies that the magnetic field $\vec{B}$ has both:

   • A vertical (axial) component: $B_z(r, z, t) = B_{0z}(z) \cos(\omega t)$
   • A horizontal (radial) component: $B_r(r, z, t) = B_{0r}(z) \cos(\omega t)$

2. Induced Current in the Ring:
   The time-varying magnetic flux $\Phi(t) = \Phi_0 \cos(\omega t)$ through the conducting ring induces an electromotive force (EMF) according to Faraday's Law:
   $\mathcal{E}(t) = -\frac{d\Phi}{dt} = \omega \Phi_0 \sin(\omega t)$

   The conducting ring has a finite electrical resistance $R$ and a self-inductance $L$. Due to this inductance, the induced current $i(t)$ lags behind the induced EMF by a phase angle $\phi$:
   $i(t) = I_{\text{ring}} \sin(\omega t - \phi)$
   where $\tan\phi = \frac{\omega L}{R}$ with $\phi > 0$.

3. Interaction and Magnetic Force:
   The magnetic Lorentz force on an element $d\vec{l} = dl\,\hat{\theta}$ of the ring carrying current $i(t)$ is given by:
   $d\vec{F} = i(t) (d\vec{l} \times \vec{B})$
   Expanding the cross product:
   $d\vec{l} \times \vec{B} = (dl\,\hat{\theta}) \times (B_r\,\hat{r} + B_z\,\hat{z}) = -B_r dl\,\hat{z} + B_z dl\,\hat{r}$

   • The radial force components cancel symmetrically around the ring or produce a net radial pinch.
   • The vertical force on the entire ring of radius $r$ is:
     $F_z(t) = -2\pi r B_r(t) i(t)$

   Taking the time average of the vertical force over a cycle:
   $\langle F_z \rangle \propto -\langle B_r(t) i(t) \rangle = -\langle B_{0r} \cos(\omega t) \cdot I_{\text{ring}} \sin(\omega t - \phi) \rangle$

   Using the identity $\sin(\omega t - \phi) = \sin(\omega t)\cos\phi - \cos(\omega t)\sin\phi$:
   $\langle \cos(\omega t)\sin(\omega t - \phi) \rangle = -\frac{1}{2}\sin\phi$

   Therefore:
   $\langle F_z \rangle \propto \frac{1}{2} B_{0r} I_{\text{ring}} \sin\phi$

   Since $L > 0$, the phase angle $\phi > 0$, making $\sin\phi > 0$. This ensures a non-zero, time-averaged magnetic force directed upward ($\langle F_z \rangle > 0$).

4. Levitation:
   As the ring moves higher above the coil, the magnetic field strength decreases, which decreases $\langle F_z \rangle$. At a certain height, the upward magnetic force balances the downward gravitational force ($mg$):
   $\langle F_z \rangle = mg$
   Hence, the ring can stably float (levitate) at this height.

Therefore:

• Statement-1 is True.
• Statement-2 is True.
• Statement-2 is the correct explanation for Statement-1.

Correct Option: (A)