A solenoid of cross-sectional area `2xx 10^(-4)m^(2)` and 900 turns has `0.6A m^(2)` magnetic moment. Then the current flowing through it is

Options

A. 2.4A
B. 2.34mA
C. 3.33A
D. 3.33mA

Explanation

To find the current flowing through the solenoid, we can use the formula for the magnetic moment (M) of a solenoid, which is given by:

\[ M = nIA \]

Where:
- \( M \) = magnetic moment (in A·m²)
- \( n \) = number of turns per unit length (in turns/m)
- \( I \) = current (in A)
- \( A \) = cross-sectional area (in m²)

Given:
- Magnetic moment \( M = 0.6 \, \text{A·m}^2 \)
- Number of turns \( N = 900 \)
- Cross-sectional area \( A = 2 \times 10^{-4} \, \text{m}^2 \)

### Step 1: Calculate the number of turns per unit length (n)
Since the number of turns \( N \) is given, we can directly use it in our calculations. However, we need to rearrange the formula to find the current \( I \).

### Step 2: Rearranging the formula for current
From the magnetic moment formula, we can express the current \( I \) as:

\[ I = \frac{M}{nA} \]

### Step 3: Substitute the values into the equation
We need to substitute the values of \( M \), \( N \), and \( A \) into the equation. Here, \( n \) can be considered as the total number of turns \( N \) since we are not given the length of the solenoid.

Thus, we can write:

\[ I = \frac{0.6}{900 \times (2 \times 10^{-4})} \]

### Step 4: Calculate the current
Now, we can perform the calculation:

1. Calculate the denominator:
\[ 900 \times (2 \times 10^{-4}) = 900 \times 0.0002 = 0.18 \]

2. Now calculate \( I \):
\[ I = \frac{0.6}{0.18} \]

3. Performing the division:
\[ I = 3.33 \, \text{A} \]

Thus, the current flowing through the solenoid is \( 3.33 \, \text{A} \).

### Final Answer:
The current flowing through the solenoid is \( 3.33 \, \text{A} \).

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