
Concept:
The direction of induced current in a coil is determined by Lenz’s law. According to Lenz’s law: The induced current flows in such a direction that it opposes the change in magnetic flux producing it. Key ideas used:
Approaching coils increase magnetic flux
Receding coils decrease magnetic flux
The induced current always opposes the {change} in flux, not the flux itself
Step 1: Effect of coil \(L_1\) on coil \(L_2\). The current in \(L_1\) is anticlockwise. If \(L_1\) is moved towards \(L_2\), the magnetic flux through \(L_2\) due to \(L_1\) increases. To oppose this increase, the induced current in \(L_2\) must produce a magnetic field in the opposite direction, which corresponds to a clockwise current.
Step 2: Effect of coil \(L_3\) on coil \(L_2\). The current in \(L_3\) is clockwise. If \(L_3\) is moved away from \(L_2\), the magnetic flux through \(L_2\) due to \(L_3\) decreases. To oppose the decrease in flux, the induced current in \(L_2\) must try to maintain the original flux direction, again requiring a clockwise current.
Step 3: Combine both effects. Both actions:
Moving \(L_1\) towards \(L_2\)
Moving \(L_3\) away from \(L_2\) produce induced currents in \(L_2\) in the same (clockwise) direction. Hence, this combination ensures that the current in the second coil is clockwise. \[ \boxed{\text{Correct option is (1)}} \]

Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]
A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_0\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _______. 