91Ó°ÊÓ

In order to find the EMF generated in the loop, we use Faraday's law which states that the EMF is equal to the negative rate of change of magnetic flux in the loop. The magnetic flux through the loop is given by the product of the magnetic field strength, the length of the conductor, and the displacement of the rod: \(\Phi = B\ell x\), where \(x\) is the position of the rod. Therefore, the generated EMF can be expressed as: \[\epsilon = -\frac{d\Phi}{dt} = -B\ell\frac{dx}{dt}\]

Since there's no resistance in the loop, the EMF is equal to the product of the loop's inductance and the rate of change of the current in the loop: \[\epsilon = L\frac{di}{dt}\] By substituting the expression for EMF in Step 1, we get: \[-B\ell\frac{dx}{dt} = L\frac{di}{dt}\]

The magnetic force on the rod is given by the product of the magnetic field strength, the current in the loop, and the length of the conductor: \[F_{B} = B\ell i\] The force will be equal to the product of the mass of the rod and its acceleration: \[F_{B} = ma\] Using Newton's second law of motion and substituting the expression for force, we get: \[B\ell i = ma\]

Rearrange the equation from Step 2 to obtain the expression for current and substitute it into the equation from Step 3: \[B\ell \left(\frac{1}{L}\int -B\ell \frac{dx}{dt} dt\right) = ma\] At this point, we can simplify the equation and perform the integration and derivation to find the expression for acceleration and obtain the time period of oscillation for the rod. Solving the above equation, we get: \[-\frac{B^{2} \ell^{2}}{L} x = ma\] Divide both sides by mass, we obtain: \[-\frac{B^{2} \ell^{2}}{mL} x = a\] Now, we recall that the relation between acceleration and displacement is given by \(\omega^2 x\), where \(\omega\) is the angular frequency of oscillation. Comparing the two equations, we have: \[\omega^{2} = \frac{B^{2} \ell^{2}}{mL}\] We can use the relation between angular frequency and time period (\(T = \frac{2\pi}{\omega}\)) to find the time period for oscillation:\[T = 2\pi \sqrt{\frac{mL}{B^{2} \ell^{2}}}\]

Use the given values \(m = 8 kg\), \(L = 2 H\), \(B = \pi T \), and \(\ell = 2 m\) into the expression for the time period and obtain the time period of oscillation: \[T = 2\pi \sqrt{\frac{(8)(2)}{(\pi)^{2} (2)^{2}}} = 2\pi\sqrt{\frac{8}{\pi^{2}}}\] After calculating the above expression, we get the time period of oscillation for the rod: \[T \approx 3.0 s\] Thus, the time period of oscillation of the rod is approximately 3 seconds.