By a change of current from 5 A to 10 A in 0.1 s, the self induced emf is 10 V. The change in the energy of the magnetic field of a coil will be

Options

A. 5J
B. 6J
C. `7.5 J`
D. 9J

Answer

To solve the problem of finding the change in the energy of the magnetic field of a coil when the current changes from 5 A to 10 A, we can follow these steps:

### Step 1: Identify the given values
- Initial current, \( I_1 = 5 \, \text{A} \)
- Final current, \( I_2 = 10 \, \text{A} \)
- Self-induced emf, \( \mathcal{E} = 10 \, \text{V} \)
- Time interval, \( \Delta t = 0.1 \, \text{s} \)

### Step 2: Use the formula for self-induced emf
The self-induced emf is given by the formula:
\[ \mathcal{E} = L \frac{di}{dt} \]
Where:
- \( L \) is the self-inductance
- \( \frac{di}{dt} \) is the rate of change of current

### Step 3: Calculate the rate of change of current
The change in current \( di \) is:
\[ di = I_2 - I_1 = 10 \, \text{A} - 5 \, \text{A} = 5 \, \text{A} \]
The rate of change of current \( \frac{di}{dt} \) is:
\[ \frac{di}{dt} = \frac{di}{\Delta t} = \frac{5 \, \text{A}}{0.1 \, \text{s}} = 50 \, \text{A/s} \]

### Step 4: Calculate the self-inductance \( L \)
Rearranging the formula for self-induced emf gives:
\[ L = \frac{\mathcal{E}}{\frac{di}{dt}} = \frac{10 \, \text{V}}{50 \, \text{A/s}} = 0.2 \, \text{H} \]

### Step 5: Calculate the change in energy
The change in energy \( \Delta U \) in the magnetic field of the coil is given by:
\[ \Delta U = \frac{1}{2} L I_2^2 - \frac{1}{2} L I_1^2 \]
Substituting the values:
\[ \Delta U = \frac{1}{2} \times 0.2 \, \text{H} \times (10 \, \text{A})^2 - \frac{1}{2} \times 0.2 \, \text{H} \times (5 \, \text{A})^2 \]
Calculating each term:
\[ \Delta U = \frac{1}{2} \times 0.2 \times 100 - \frac{1}{2} \times 0.2 \times 25 \]
\[ = 0.1 \times 100 - 0.1 \times 25 \]
\[ = 10 - 2.5 = 7.5 \, \text{J} \]

### Final Answer
The change in the energy of the magnetic field of the coil is \( 7.5 \, \text{J} \).

---