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 find the amount of current required to demagnetize the small bar magnet when inserted into a solenoid, we can follow these steps:

### Step 1: Understand the Coercivity
The coercivity (H) of the bar magnet is given as \(4 \times 10^3 \, \text{A/m}\). This is the magnetic field strength required to demagnetize the magnet.

### Step 2: Determine the Parameters of the Solenoid
The solenoid has:
- Number of turns (N) = 500
- Length (L) = 1 m

### Step 3: Calculate the Number of Turns per Unit Length
The number of turns per unit length (n) can be calculated as:
\[ n = \frac{N}{L} = \frac{500}{1} = 500 \, \text{turns/m} \]

### Step 4: Relate the Magnetic Field to Current
The magnetic field strength (H) inside the solenoid is related to the current (I) flowing through it by the formula:
\[ H = n \cdot I \]

### Step 5: Solve for Current (I)
We can rearrange the formula to find the current:
\[ I = \frac{H}{n} \]
Substituting the known values:
\[ I = \frac{4 \times 10^3 \, \text{A/m}}{500 \, \text{turns/m}} = \frac{4 \times 10^3}{500} \]

### Step 6: Simplify the Calculation
Calculating the right-hand side:
\[ I = \frac{4 \times 10^3}{500} = \frac{4000}{500} = 8 \, \text{A} \]

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

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