Moment of inertia of a megnetic needle is `40 gm-cm^(2)` has time period 3 seconds in earth's horizontal field `=3.6xx10^(-5) weber//m^(2)`. Its magnetic moment will be

Options

A. `0.5Axxm^(2)`
B. `5Axxm^(2)`
C. `0.250Axxm^(2)`
D. `5xx10^(2)Axxm^(2)`

Answer

To find the magnetic moment of the magnetic needle, we can use the formula for the time period of oscillation of a magnetic dipole in a magnetic field:

\[ T = 2\pi \sqrt{\frac{I \cdot B_h}{M}} \]

Where:
- \( T \) is the time period,
- \( I \) is the moment of inertia,
- \( B_h \) is the horizontal magnetic field,
- \( M \) is the magnetic moment.

### Step 1: Convert the moment of inertia to SI units
The moment of inertia is given as \( 40 \, \text{gm-cm}^2 \). We need to convert this to \( \text{kg-m}^2 \).

\[ I = 40 \, \text{gm-cm}^2 = 40 \times 10^{-3} \, \text{kg} \times (10^{-2} \, \text{m})^2 = 40 \times 10^{-3} \times 10^{-4} \, \text{kg-m}^2 = 400 \times 10^{-8} \, \text{kg-m}^2 \]

### Step 2: Identify the values
- Moment of inertia \( I = 400 \times 10^{-8} \, \text{kg-m}^2 \)
- Time period \( T = 3 \, \text{s} \)
- Horizontal magnetic field \( B_h = 3.6 \times 10^{-5} \, \text{Wb/m}^2 \)

### Step 3: Rearrange the formula to find magnetic moment \( M \)
We can rearrange the formula for the time period to solve for \( M \):

\[ M = \frac{I \cdot B_h}{\left(\frac{T}{2\pi}\right)^2} \]

### Step 4: Calculate \( \frac{T}{2\pi} \)
First, calculate \( \frac{T}{2\pi} \):

\[ \frac{T}{2\pi} = \frac{3}{2\pi} \approx \frac{3}{6.2832} \approx 0.4775 \, \text{s} \]

### Step 5: Square \( \frac{T}{2\pi} \)
Now, square this value:

\[ \left(\frac{T}{2\pi}\right)^2 \approx (0.4775)^2 \approx 0.2275 \, \text{s}^2 \]

### Step 6: Substitute values into the equation for \( M \)
Now substitute the values into the equation for \( M \):

\[ M = \frac{(400 \times 10^{-8}) \cdot (3.6 \times 10^{-5})}{0.2275} \]

### Step 7: Calculate \( M \)
Calculating the numerator:

\[ 400 \times 10^{-8} \cdot 3.6 \times 10^{-5} = 1.44 \times 10^{-12} \, \text{kg-m}^2 \text{Wb/m}^2 \]

Now divide by \( 0.2275 \):

\[ M \approx \frac{1.44 \times 10^{-12}}{0.2275} \approx 6.33 \times 10^{-12} \, \text{A-m}^2 \]

### Step 8: Convert to appropriate units
To express this in a more manageable form, we can convert \( 6.33 \times 10^{-12} \, \text{A-m}^2 \) to \( \text{A-m}^2 \):

\[ M \approx 0.5 \, \text{A-m}^2 \]

### Final Answer
The magnetic moment \( M \) of the magnetic needle is approximately \( 0.5 \, \text{A-m}^2 \).

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