A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The coercivity of the bar magnet is :

Explanation

To find the coercivity of the bar magnet, we will follow these steps:

### Step 1: Understand the relationship between magnetic field and coercivity.
The magnetic field \( B \) inside a solenoid can be expressed using the formula:
\[ B = \mu_0 H \]
where \( \mu_0 \) is the permeability of free space, and \( H \) is the magnetic field strength (which is also referred to as coercivity in this context).

### Step 2: Use the formula for the magnetic field inside a solenoid.
The magnetic field \( B \) inside a solenoid can also be expressed as:
\[ B = \mu_0 n I \]
where \( n \) is the number of turns per unit length of the solenoid, and \( I \) is the current flowing through the solenoid.

### Step 3: Relate the two expressions for magnetic field.
By equating the two expressions for \( B \):
\[ \mu_0 H = \mu_0 n I \]
We can simplify this by dividing both sides by \( \mu_0 \) (assuming \( \mu_0 \neq 0 \)):
\[ H = n I \]

### Step 4: Calculate the number of turns per unit length \( n \).
The number of turns \( N \) is given as 100, and the length \( l \) of the solenoid is 0.2 m. Therefore, the number of turns per unit length \( n \) is:
\[ n = \frac{N}{l} = \frac{100}{0.2} = 500 \, \text{turns/m} \]

### Step 5: Substitute the values into the coercivity formula.
Now, we can substitute \( n \) and the current \( I = 5.2 \, \text{A} \) into the formula for \( H \):
\[ H = n I = 500 \times 5.2 \]

### Step 6: Perform the calculation.
Calculating the above expression:
\[ H = 500 \times 5.2 = 2600 \, \text{A/m} \]

### Conclusion:
Thus, the coercivity of the bar magnet is:
\[ \text{Coercivity} = 2600 \, \text{A/m} \]