A short bar magnet is placed in the magnetic meridian of the earth with its north pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East-West line drawn through the mid point of the magnet. What is the magnetic moment of the magnet in `Am^(2)`? (Given `m=10^(-7)` in SJ units and `B_(H)=` horizontal component of earth's magnetic field `=3.6xx10^(-5)` tesla)

Options

A. 14.6
B. 19.4
C. 9.7
D. 4.9

Explanation

To find the magnetic moment of the bar magnet, we can follow these steps:

### Step 1: Understand the setup
We have a short bar magnet placed in the magnetic meridian of the Earth with its north pole pointing north. The neutral points are located 30 cm away from the magnet on the East-West line through its midpoint.

### Step 2: Identify the given values
- Distance to neutral points, \( r = 30 \, \text{cm} = 0.30 \, \text{m} \)
- Horizontal component of Earth's magnetic field, \( B_H = 3.6 \times 10^{-5} \, \text{T} \)
- Magnetic permeability in SI units, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)

### Step 3: Use the formula for the magnetic field due to a bar magnet
At the neutral point, the magnetic field due to the bar magnet is equal in magnitude and opposite in direction to the horizontal component of the Earth's magnetic field. The formula for the magnetic field \( B \) at a distance \( r \) from a magnetic dipole (bar magnet) is given by:

\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3} \]

Where \( m \) is the magnetic moment of the magnet.

### Step 4: Set up the equation
At the neutral point, we have:

\[ B_H = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3} \]

### Step 5: Substitute the known values into the equation
Substituting \( B_H \), \( \mu_0 \), and \( r \):

\[ 3.6 \times 10^{-5} = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2m}{(0.30)^3} \]

This simplifies to:

\[ 3.6 \times 10^{-5} = \frac{10^{-7} \cdot 2m}{0.027} \]

### Step 6: Solve for \( m \)
Rearranging gives:

\[ m = \frac{3.6 \times 10^{-5} \cdot 0.027}{2 \times 10^{-7}} \]

Calculating the right-hand side:

\[ m = \frac{9.72 \times 10^{-7}}{2 \times 10^{-7}} = 4.86 \, \text{A m}^2 \]

### Step 7: Final result
Thus, the magnetic moment of the magnet is:

\[ m \approx 4.86 \, \text{A m}^2 \]