The current through a solenoid when it replaces a bar magnet `A = 2 xx 10^(-4) m^(2)` and turns are 1000 but same magnetic moment `(0.04) Am^(2)`:

Options

A. 4A
B. 2A
C. 3A
D. `1 A`

Explanation

To find the current through the solenoid that replaces a bar magnet, we can use the formula for the magnetic moment \( m \) of a solenoid, which is given by:

\[ m = n \cdot I \cdot A \]

Where:
- \( m \) is the magnetic moment,
- \( n \) is the number of turns per unit length,
- \( I \) is the current,
- \( A \) is the cross-sectional area of the solenoid.

Given:
- \( m = 0.04 \, \text{Am}^2 \)
- \( A = 2 \times 10^{-4} \, \text{m}^2 \)
- Total number of turns \( N = 1000 \)

### Step 1: Calculate \( n \)
Since the number of turns \( N \) is given, we can find \( n \) as:

\[ n = \frac{N}{L} \]

However, we do not have the length \( L \) of the solenoid. For this problem, we can assume that we are using the total number of turns directly in the formula without dividing by length, as we are looking for the current directly.

### Step 2: Rearrange the formula to solve for \( I \)
We can rearrange the formula for magnetic moment to solve for current \( I \):

\[ I = \frac{m}{n \cdot A} \]

### Step 3: Substitute the values into the equation
Now we can substitute the known values into the equation:

\[ I = \frac{0.04}{1000 \cdot (2 \times 10^{-4})} \]

### Step 4: Calculate the denominator
Calculate \( 1000 \cdot (2 \times 10^{-4}) \):

\[ 1000 \cdot (2 \times 10^{-4}) = 2 \times 10^{-1} = 0.2 \]

### Step 5: Substitute back into the equation for \( I \)
Now substitute this back into the equation for \( I \):

\[ I = \frac{0.04}{0.2} \]

### Step 6: Perform the division
Now perform the division:

\[ I = 0.2 \, \text{A} \]

### Final Answer
The current through the solenoid is:

\[ I = 0.2 \, \text{A} \]

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