Find the energy stored in the magnetic field if current of 5A produces a magnetic flux of `2xx10^(-3)` Wb through a coil of 500 turns.

Options

A. 2.5J
B. 0.25J
C. 250J
D. 1.5J

Answer

To find the energy stored in the magnetic field of a coil, we can use the formula for energy stored in an inductor, which is given by:

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

Where:
- \( U \) is the energy stored in joules,
- \( L \) is the inductance in henries,
- \( I \) is the current in amperes.

### Step 1: Calculate the Inductance \( L \)

The inductance \( L \) can be calculated using the formula:

\[ L = \frac{N \Phi}{I} \]

Where:
- \( N \) is the number of turns in the coil,
- \( \Phi \) is the magnetic flux in webers,
- \( I \) is the current in amperes.

Given:
- \( N = 500 \) turns,
- \( \Phi = 2 \times 10^{-3} \) Wb,
- \( I = 5 \) A.

Substituting the values:

\[ L = \frac{500 \times (2 \times 10^{-3})}{5} \]

Calculating this gives:

\[ L = \frac{1000 \times 10^{-3}}{5} = \frac{1}{5} = 0.2 \, \text{H} \]

### Step 2: Calculate the Energy Stored \( U \)

Now that we have the inductance \( L \), we can substitute it back into the energy formula:

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

Substituting \( L = 0.2 \, \text{H} \) and \( I = 5 \, \text{A} \):

\[ U = \frac{1}{2} \times 0.2 \times (5)^2 \]

Calculating this gives:

\[ U = \frac{1}{2} \times 0.2 \times 25 = 0.1 \times 25 = 2.5 \, \text{J} \]

### Final Answer

The energy stored in the magnetic field is:

\[ \boxed{2.5 \, \text{J}} \]