A coil has an inductance of `50 m H` and a resistance of `0.3Omega`. If a `12V` emf is applied across the coil, the energy stored in the magnetic field after the current has built up of its steady state value is

Options

A. `40J`
B. `40mJ`
C. `20J`
D. `20mJ`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Inductance (L) = 50 mH = \(50 \times 10^{-3}\) H
- Resistance (R) = 0.3 Ω
- EMF (E) = 12 V

### Step 2: Calculate the steady-state current (I)
Using Ohm's Law, the current (I) in the circuit can be calculated as:
\[ I = \frac{E}{R} \]
Substituting the values:
\[ I = \frac{12 \text{ V}}{0.3 \text{ Ω}} = 40 \text{ A} \]

### Step 3: Calculate the energy stored in the magnetic field
The energy (U) stored in the magnetic field of an inductor is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
Substituting the values we have:
\[ U = \frac{1}{2} \times (50 \times 10^{-3} \text{ H}) \times (40 \text{ A})^2 \]
Calculating \(I^2\):
\[ (40 \text{ A})^2 = 1600 \text{ A}^2 \]
Now substituting this back into the energy formula:
\[ U = \frac{1}{2} \times (50 \times 10^{-3}) \times 1600 \]
Calculating further:
\[ U = \frac{1}{2} \times 50 \times 1600 \times 10^{-3} \]
\[ U = 25 \times 1600 \times 10^{-3} \]
\[ U = 40000 \times 10^{-3} = 40 \text{ J} \]

### Final Answer
The energy stored in the magnetic field after the current has built up to its steady state value is **40 Joules**.

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