The self inductance and resistance of coil are 5H and `20 Omega` respectively . On applying an e.m.f . Of 100 V on it , the magnetic potential energy stored in the coil will be __

Options

A. 62.5 Joule
B. 62.5 erg
C. `62.5xx10^(-3)` Joule
D. zero

Answer

To find the magnetic potential energy stored in the coil, we can follow these steps:

### Step 1: Identify the given values
- Self-inductance (L) = 5 H
- Resistance (R) = 20 Ω
- EMF (V) = 100 V

### Step 2: Calculate the steady-state current (I)
In a steady state, the inductor behaves like a short circuit. Therefore, the current can be calculated using Ohm's law:

\[ I = \frac{V}{R} \]

Substituting the values:

\[ I = \frac{100 \, \text{V}}{20 \, \Omega} = 5 \, \text{A} \]

### Step 3: Calculate the magnetic potential energy (U)
The magnetic potential energy stored in the inductor is given by the formula:

\[ U = \frac{1}{2} L I^2 \]

Substituting the values of L and I:

\[ U = \frac{1}{2} \times 5 \, \text{H} \times (5 \, \text{A})^2 \]

Calculating \(I^2\):

\[ I^2 = 5^2 = 25 \]

Now substituting back into the energy formula:

\[ U = \frac{1}{2} \times 5 \times 25 = \frac{125}{2} = 62.5 \, \text{J} \]

### Final Answer
The magnetic potential energy stored in the coil is **62.5 J**.

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