A bar magnet 8 cm long is placed in the magnetic merdian with the N-pole pointing towards geographical north . Two netural points separated by a distance of 6 cms are obtained on the equatorial axis of the magnet . If horizontal component of earth's field `=3.2xx10^(-5)T,` then pole strength of magnet is

Options

A. `5 ab-ampxxcm`
B. `10 ab-amp xx cm`
C. `2.5 ab- amp xx cm`
D. `20 ab- ampxx cm`

Explanation

To solve the problem of finding the pole strength of the bar magnet, we can follow these steps:

### Step 1: Identify the Given Data
- Length of the bar magnet (2l) = 8 cm, hence l = 4 cm = 0.04 m
- Distance between the two neutral points (d) = 6 cm, hence d = 3 cm = 0.03 m
- Horizontal component of Earth's magnetic field (H) = \(3.2 \times 10^{-5}\) T

### Step 2: Use the Formula for the Magnetic Field on the Equatorial Axis
The magnetic field \(H\) at a point on the equatorial axis of a magnetic dipole is given by the formula:
\[ H = \frac{\mu_0}{4\pi} \cdot \frac{2M}{(d^2 + l^2)^{3/2}} \]
where \(M\) is the magnetic moment of the magnet and \(\mu_0\) is the permeability of free space, approximately \(4\pi \times 10^{-7} \, \text{T m/A}\).

### Step 3: Rearrange the Formula to Find Magnetic Moment \(M\)
Rearranging the formula gives:
\[ M = \frac{H \cdot (d^2 + l^2)^{3/2} \cdot 4\pi}{2\mu_0} \]

### Step 4: Calculate \(d^2 + l^2\)
Calculate \(d^2 + l^2\):
\[ d^2 = (0.03)^2 = 0.0009 \, \text{m}^2 \]
\[ l^2 = (0.04)^2 = 0.0016 \, \text{m}^2 \]
\[ d^2 + l^2 = 0.0009 + 0.0016 = 0.0025 \, \text{m}^2 \]

### Step 5: Calculate \((d^2 + l^2)^{3/2}\)
\[ (d^2 + l^2)^{3/2} = (0.0025)^{3/2} = (0.0025)^{1.5} = 0.000125 \, \text{m}^3 \]

### Step 6: Substitute Values into the Magnetic Moment Formula
Substituting the values into the rearranged formula:
\[ M = \frac{(3.2 \times 10^{-5}) \cdot (0.000125) \cdot 4\pi}{2 \cdot (4\pi \times 10^{-7})} \]
The \(4\pi\) cancels out:
\[ M = \frac{(3.2 \times 10^{-5}) \cdot (0.000125)}{2 \cdot (10^{-7})} \]
Calculating this gives:
\[ M = \frac{(3.2 \times 10^{-5}) \cdot (0.000125)}{2 \cdot (10^{-7})} = \frac{4.0 \times 10^{-9}}{2 \times 10^{-7}} = 2.0 \times 10^{-2} \, \text{A m}^2 \]

### Step 7: Calculate the Pole Strength \(m\)
The pole strength \(m\) can be calculated using the formula:
\[ m = \frac{M}{2l} \]
Substituting \(M\) and \(l\):
\[ m = \frac{2.0 \times 10^{-2}}{2 \cdot 0.04} = \frac{2.0 \times 10^{-2}}{0.08} = 0.25 \, \text{A m} \]

### Step 8: Convert to A cm
To convert to A cm:
\[ m = 0.25 \, \text{A m} = 25 \, \text{A cm} \]

### Final Answer
The pole strength of the magnet is \(25 \, \text{A cm}\).

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