A short bar magnet has a magnetic moment of `0.65TJ ^(-1)`, then the magnitude and direction of the magnetic field produced by the magnet at a distance 8 cm from the centre of magnet on the axis is

Options

A. `2.5 xx 10^(-4)T`, along NS direction
B. `2.5 xx 10^(-4)T`, along SN direction
C. `4.5 xx 10^(-4)T`, along NS direction
D. `4.5 xx 10^(-4)T`, along NS direction

Explanation

To find the magnitude and direction of the magnetic field produced by a short bar magnet at a distance of 8 cm from its center on the axis, we can follow these steps:

### Step 1: Understand the magnetic moment
The magnetic moment (m) of the bar magnet is given as \(0.65 \, \text{T J}^{-1}\). This value represents the strength of the magnet.

### Step 2: Identify the formula for the magnetic field
For a short bar magnet, the magnetic field (B) at a distance (d) from the center of the magnet along its axis can be calculated using the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} \]
where:
- \(\mu_0\) is the permeability of free space, approximately \(4\pi \times 10^{-7} \, \text{T m/A}\),
- \(m\) is the magnetic moment,
- \(d\) is the distance from the center of the magnet.

### Step 3: Convert the distance to meters
The distance given is 8 cm. We need to convert this to meters:
\[ d = 8 \, \text{cm} = 0.08 \, \text{m} \]

### Step 4: Substitute the values into the formula
Now, substituting the values into the formula:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 0.65}{(0.08)^3} \]
This simplifies to:
\[ B = 10^{-7} \cdot \frac{2 \times 0.65}{(0.08)^3} \]

### Step 5: Calculate \(d^3\)
Calculating \(d^3\):
\[ (0.08)^3 = 0.000512 \, \text{m}^3 \]

### Step 6: Substitute \(d^3\) back into the equation
Now substituting \(d^3\) back into the equation:
\[ B = 10^{-7} \cdot \frac{1.3}{0.000512} \]

### Step 7: Calculate the value
Calculating the fraction:
\[ \frac{1.3}{0.000512} \approx 2539.0625 \]
Now substituting this back:
\[ B \approx 10^{-7} \cdot 2539.0625 \approx 2.539 \times 10^{-4} \, \text{T} \]

### Step 8: Round off the value
Rounding off, we get:
\[ B \approx 2.5 \times 10^{-4} \, \text{T} \]

### Step 9: Determine the direction
The direction of the magnetic field produced by the bar magnet is from the North pole to the South pole. Since we are measuring along the axis of the magnet, the direction will be from South to North.

### Final Answer
The magnitude of the magnetic field at a distance of 8 cm from the center of the magnet on the axis is \(2.5 \times 10^{-4} \, \text{T}\) along the South-North direction.

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