A coil has an inductance of $2 \mathrm{H}$ and resistance of $4 ~\Omega$. A $10 \mathrm{~V}$ is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ___________ $\times 10^{-2} \mathrm{~J}$.

<p>To find the energy stored in the magnetic field after the current has built up to its equilibrium value, we first need to find the steady-state current in the coil. </p> <p>When the current reaches its equilibrium value, the coil behaves like a resistor because the back-emf induced by the changing magnetic field is zero. Ohm&#39;s law can be applied:</p> <p>$I = \frac{V}{R}$</p> <p>where</p> <ul> <li>$I$ is the current</li> <li>$V$ is the voltage across the coil (10 V)</li> <li>$R$ is the resistance of the coil (4 Ω)</li> </ul> <p>Plugging in the values:</p> <p>$I = \frac{10}{4}$ $I = 2.5 \mathrm{~A}$</p> <p>Now that we have the steady-state current, we can find the energy stored in the magnetic field using the formula:</p> <p>$W = \frac{1}{2}LI^2$</p> <p>where</p> <ul> <li>$W$ is the energy stored in the magnetic field</li> <li>$L$ is the inductance of the coil (2 H)</li> <li>$I$ is the steady-state current (2.5 A)</li> </ul> <p>Plugging in the values:</p> <p>$W = \frac{1}{2}(2)(2.5)^2$<br/><br/> $W = 1(6.25)$<br/><br/> $W = 6.25 \mathrm{~J}$</p> <p>To express this in terms of $10^{-2} \mathrm{~J}$, divide by $10^{-2}$:</p> <p>$6.25 \div 10^{-2} = 625$</p> <p>Therefore, the energy stored in the magnetic field after the current has built up to its equilibrium value is 625$\times 10^{-2} \mathrm{~J}$.</p>