QJEEM23Phe9c1: JEE Main 2023 Physics - Class 12 JEEM23Phe9c1 Solution: A coil has an inductance of 2 H and resistance of 4   Ω .…

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$$\times {10}^{-2}\mathrm{J}$$.