JEE Main 2023 — Alternating Current Question with Solution

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:

$$W=\frac{1}{2}L{I}^2$$

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 $$10^{-2}$$:

$$6.25÷10^{-2}=625$$

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