The coercivity of a small bar magnet is `4xx10^3Am^(-1)` . It is inserted inside a solenoid of 500 turns and length 1 m to demagnetize it. The amount of current to be passed through the solenoid will be

Options

A. 2.5 A
B. `5A`
C. `8A`
D. `10A`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
- Coercivity (H) of the magnet = \(4 \times 10^3 \, \text{A/m}\)
- Number of turns (N) of the solenoid = 500 turns
- Length (L) of the solenoid = 1 m

### Step 2: Calculate the Number of Turns per Unit Length
Since the length of the solenoid is 1 m, the number of turns per unit length (n) is equal to the total number of turns:
\[ n = N / L = 500 / 1 = 500 \, \text{turns/m} \]

### Step 3: Use the Formula to Find Current
The relationship between magnetic intensity (H), number of turns per unit length (n), and current (I) is given by:
\[ I = \frac{H}{n} \]
Substituting the values we have:
\[ I = \frac{4 \times 10^3 \, \text{A/m}}{500 \, \text{turns/m}} \]

### Step 4: Perform the Calculation
Calculating the current:
\[ I = \frac{4000}{500} = 8 \, \text{A} \]

### Conclusion
The amount of current to be passed through the solenoid to demagnetize the magnet is \(8 \, \text{A}\).

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