The self-inductance of a coil is \(20 \mathrm{mH}\). When a current flows in it, one-fifth of its magnetic flux links a second coil. What electromotance is induced in the second coil if a current in the first collapses at a uniform rate of \(0.5 \mathrm{~A} \mathrm{~s}^{-1} ?\)

#tag_title#Step 2: Calculate the magnetic flux in the first coil #tag_content#To calculate the magnetic flux in the first coil, we can use the formula: \(Φ_1 = L_1 × i_1\) where: \(Φ_1\) is the magnetic flux in the first coil, \(L_1\) is the self-inductance of the first coil, and \(i_1\) is the rate of current decrease in the first coil. We are given \(L_1 = 20 \mathrm{mH}\) and \(i_1 = 0.5 \mathrm{~A~s^{-1}}\). Plugging these values into the formula, we get: \(Φ_1 = (20 \times 10^{-3} \mathrm{H}) × (0.5 \mathrm{A~s^{-1}}) = 10 × 10^{-3} \mathrm{Wb}\) So, the magnetic flux in the first coil is \(10 \mathrm{mWb}\). #tag_title#Step 3: Calculate the magnetic flux linked to the second coil #tag_content#To find the magnetic flux linked to the second coil, we can calculate the fraction of the magnetic flux in the first coil that links to the second coil using the ratio \(k\). We are given \(k = 1/5\). Multiplying the magnetic flux in the first coil by the ratio, we get: \(Φ_2 = k × Φ_1\) \(Φ_2 = (1/5) × (10 × 10^{-3} \mathrm{Wb}) = 2 × 10^{-3} \mathrm{Wb}\) So, the magnetic flux linked to the second coil is \(2 \mathrm{mWb}\). #tag_title#Step 4: Calculate the induced emf in the second coil #tag_content#Finally, we can use Faraday's law of electromagnetic induction to find the induced emf in the second coil. Faraday's law states that: \(emf = - \dfrac{dΦ}{dt}\) where: \(emf\) is the induced electromotive force in the second coil, and \(\dfrac{dΦ}{dt}\) is the rate of change of magnetic flux in the second coil. We have \(Φ_2 = 2 \mathrm{mWb}\), and since the rate of decrease in the current is constant (\(i_1\) is constant), the rate of change of magnetic flux will be the same for the second coil. Therefore, we can use the rate of change of magnetic flux in the first coil for the second coil as well. \(\dfrac{dΦ}{dt} = 10\times10^{-3} \mathrm{Wb~s^{-1}}\) Now, we can use Faraday's law to find the induced emf in the second coil: \(emf = - \dfrac{dΦ}{dt} = - (2 × 10^{-3} \mathrm{Wb~s^{-1}}) = -2 \mathrm{mV}\) Since the negative sign indicates the direction of the emf, we can say that the induced electromotive force in the second coil is \(2 \mathrm{mV}\).