The coercivity of a bar magnet is `120A//m`. It is to be demagnetised by placing it inside a solenoid of length `120cm` and number of turns 72. The current (in A) flowing through the solenoid is:

Options

A. 4A
B. 2A
C. 1A
D. zero

Answer

To solve the problem, we need to find the current flowing through a solenoid that can demagnetize a bar magnet with a coercivity of 120 A/m. The solenoid has a length of 120 cm and 72 turns.

### Step-by-Step Solution:

1. **Convert the Length of the Solenoid to Meters:**
The length of the solenoid is given as 120 cm. To convert this to meters:
\[ L = 120 \, \text{cm} = \frac{120}{100} \, \text{m} = 1.2 \, \text{m} \]

2. **Calculate the Number of Turns per Unit Length (n):**
The number of turns per unit length \( n \) is calculated using the formula:
\[ n = \frac{N}{L} \]
where \( N \) is the total number of turns and \( L \) is the length of the solenoid.
\[ n = \frac{72}{1.2} = 60 \, \text{turns/m} \]

3. **Use the Formula for Magnetic Field Strength (H):**
The magnetic field strength \( H \) inside a solenoid is given by the formula:
\[ H = n \cdot I \]
where \( I \) is the current flowing through the solenoid.

4. **Set Up the Equation Using Coercivity:**
We know that the coercivity \( H \) of the bar magnet is 120 A/m. Therefore, we can set up the equation:
\[ 120 = n \cdot I \]
Substituting the value of \( n \):
\[ 120 = 60 \cdot I \]

5. **Solve for Current (I):**
Rearranging the equation to find \( I \):
\[ I = \frac{120}{60} = 2 \, \text{A} \]

### Final Answer:
The current flowing through the solenoid is \( 2 \, \text{A} \).

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