Case 1: Number of field lines through an area decreases from 100 to 99 in 10 seconds
Case II: Number of field lines through an area decreases from 10 to 0 in 1 second In which case will the induced emf be higher?

Options

A. Case I
B. Case II
C. Same in both
D. Data insufficient

Explanation

To determine in which case the induced electromotive force (emf) is higher, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced emf (ε) is given by the rate of change of magnetic flux (Φ) through a circuit:

\[ \varepsilon = -\frac{d\Phi}{dt} \]

Where:
- \( \varepsilon \) is the induced emf,
- \( d\Phi \) is the change in magnetic flux,
- \( dt \) is the time interval over which the change occurs.

### Step 1: Calculate the change in magnetic flux for both cases.

**Case 1:**
- Initial number of field lines = 100
- Final number of field lines = 99
- Change in field lines (\( d\Phi \)) = 99 - 100 = -1
- Time interval (\( dt \)) = 10 seconds

**Case 2:**
- Initial number of field lines = 10
- Final number of field lines = 0
- Change in field lines (\( d\Phi \)) = 0 - 10 = -10
- Time interval (\( dt \)) = 1 second

### Step 2: Calculate the induced emf for both cases.

**Induced emf for Case 1:**
\[ \varepsilon_1 = -\frac{d\Phi}{dt} = -\frac{-1}{10} = \frac{1}{10} \text{ V} \]

**Induced emf for Case 2:**
\[ \varepsilon_2 = -\frac{d\Phi}{dt} = -\frac{-10}{1} = 10 \text{ V} \]

### Step 3: Compare the induced emf values.

From the calculations:
- Induced emf in Case 1 (\( \varepsilon_1 \)) = 0.1 V
- Induced emf in Case 2 (\( \varepsilon_2 \)) = 10 V

### Conclusion:
The induced emf is higher in Case 2.