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.

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's law can be applied:

$$I = \frac{V}{R}$$

where

• $$I$$ is the current
• $$V$$ is the voltage across the coil (10 V)
• $$R$$ is the resistance of the coil (4 Ω)

Plugging in the values:

$$I = \frac{10}{4}$$ $$I = 2.5 \mathrm{~A}$$

Now that we have the steady-state current, we can find the energy stored in the magnetic field using the formula:

$$W = \frac{1}{2}LI^2$$

where

• $$W$$ is the energy stored in the magnetic field
• $$L$$ is the inductance of the coil (2 H)
• $$I$$ is the steady-state current (2.5 A)

Plugging in the values:

$$W = \frac{1}{2}(2)(2.5)^2$$

$$W = 1(6.25)$$

$$W = 6.25 \mathrm{~J}$$

To express this in terms of $$10^{-2} \mathrm{~J}$$, divide by $$10^{-2}$$:

$$6.25 \div 10^{-2} = 625$$

Therefore, the energy stored in the magnetic field after the current has built up to its equilibrium value is 625$$\times 10^{-2} \mathrm{~J}$$.