A short bar magnet has a magnetic moment of `0*48JT^-1`. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of `10cm` from the centre of the magnet on (i) the axis (ii) the equatorial line (normal bisector) of the magnet.

Options

A. `0.48xx10^(-4)` T along N-S direction
B. `0.28xx10^(-4)` T along S-N direction
C. `0.28xx10^(-4)` T along N-S direction
D. `0.96xx10^(-4)` T along S-N direction

Explanation

To solve the problem, we will calculate the magnetic field produced by a short bar magnet at a distance of 10 cm from its center, both on the axis and on the equatorial line.

### Given Data:
- Magnetic moment, \( m = 0.48 \, \text{JT}^{-1} \)
- Distance, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)

### (i) Magnetic Field on the Axis (BA)

1. **Formula for Magnetic Field on the Axis:**
\[ B_A = \frac{\mu_0 \cdot 2m}{4\pi r^3} \]

2. **Substituting the Values:**
\[ B_A = \frac{(4\pi \times 10^{-7}) \cdot 2 \cdot 0.48}{4\pi (0.1)^3} \]

3. **Simplifying the Equation:**
The \( 4\pi \) cancels out:
\[ B_A = \frac{2 \cdot 0.48 \times 10^{-7}}{(0.1)^3} \]
\[ B_A = \frac{0.96 \times 10^{-7}}{0.001} = 0.96 \times 10^{-4} \, \text{T} \]

4. **Final Result for Magnetic Field on the Axis:**
\[ B_A = 0.96 \times 10^{-4} \, \text{T} = 9.6 \times 10^{-5} \, \text{T} \]

### (ii) Magnetic Field on the Equatorial Line (BE)

1. **Formula for Magnetic Field on the Equatorial Line:**
\[ B_E = \frac{\mu_0 m}{4\pi r^3} \]

2. **Substituting the Values:**
\[ B_E = \frac{(4\pi \times 10^{-7}) \cdot 0.48}{4\pi (0.1)^3} \]

3. **Simplifying the Equation:**
The \( 4\pi \) cancels out:
\[ B_E = \frac{0.48 \times 10^{-7}}{(0.1)^3} \]
\[ B_E = \frac{0.48 \times 10^{-7}}{0.001} = 0.48 \times 10^{-4} \, \text{T} \]

4. **Final Result for Magnetic Field on the Equatorial Line:**
\[ B_E = 0.48 \times 10^{-4} \, \text{T} = 4.8 \times 10^{-5} \, \text{T} \]

### Direction of the Magnetic Field

- The direction of the magnetic field is always from the North Pole to the South Pole of the magnet. Therefore, at any point in the plane, the magnetic field will point from the North to the South.

### Summary of Results:
- **Magnetic Field on the Axis:** \( B_A = 9.6 \times 10^{-5} \, \text{T} \)
- **Magnetic Field on the Equatorial Line:** \( B_E = 4.8 \times 10^{-5} \, \text{T} \)
- **Direction:** From North Pole to South Pole.