A bar magnet has a pole strength of `3.6 A-m` and magnetic length `8 cm`.Find the magnetic at (a) a point on the axis at a distance of `6 cm` from the centre towards the north pole and (b) a point on the perpendicular bisector at the same distance.

Explanation

To solve the problem, we will calculate the magnetic field at two different points in relation to the bar magnet. Let's break it down step by step.

### Given Data:
- Pole strength (m) = 3.6 A-m
- Magnetic length (L) = 8 cm = 0.08 m
- Distance from the center towards the north pole (d) = 6 cm = 0.06 m

### Part (a): Magnetic Field on the Axis
1. **Identify the formula for the magnetic field on the axis of a bar magnet:**
The magnetic field (B1) at a point on the axis of the magnet is given by:
\[ B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3 - L^3} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

2. **Calculate the magnetic moment (M):**
The magnetic moment (M) is given by:
\[ M = m \cdot L = 3.6 \, \text{A-m} \cdot 0.08 \, \text{m} = 0.288 \, \text{A-m}^2 \]

3. **Substitute the values into the formula:**
\[ B_1 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \cdot 0.288}{(0.06)^3 - (0.08)^3} \]
Simplifying gives:
\[ B_1 = 10^{-7} \cdot \frac{0.576}{(0.06)^3 - (0.08)^3} \]

4. **Calculate \( (0.06)^3 \) and \( (0.08)^3 \):**
\[ (0.06)^3 = 0.000216 \quad \text{and} \quad (0.08)^3 = 0.000512 \]
Therefore,
\[ (0.06)^3 - (0.08)^3 = 0.000216 - 0.000512 = -0.000296 \]

5. **Calculate B1:**
\[ B_1 = 10^{-7} \cdot \frac{0.576}{-0.000296} \approx 8.6 \times 10^{-4} \, \text{T} \]

### Part (b): Magnetic Field on the Perpendicular Bisector
1. **Identify the formula for the magnetic field on the perpendicular bisector:**
The magnetic field (B2) at a point on the perpendicular bisector is given by:
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{m}{(d^2 + (L/2)^2)^{3/2}} \]

2. **Calculate \( L/2 \):**
\[ L/2 = 0.08/2 = 0.04 \, \text{m} \]

3. **Substitute the values into the formula:**
\[ B_2 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{3.6}{(0.06^2 + 0.04^2)^{3/2}} \]
Simplifying gives:
\[ B_2 = 10^{-7} \cdot \frac{3.6}{(0.06^2 + 0.04^2)^{3/2}} \]

4. **Calculate \( (0.06^2 + 0.04^2) \):**
\[ (0.06)^2 = 0.0036 \quad \text{and} \quad (0.04)^2 = 0.0016 \]
Therefore,
\[ 0.06^2 + 0.04^2 = 0.0036 + 0.0016 = 0.0052 \]

5. **Calculate \( (0.0052)^{3/2} \):**
\[ (0.0052)^{3/2} \approx 0.000117 \]

6. **Calculate B2:**
\[ B_2 = 10^{-7} \cdot \frac{3.6}{0.000117} \approx 7.7 \times 10^{-5} \, \text{T} \]

### Final Answers:
- (a) The magnetic field at a point on the axis is approximately \( 8.6 \times 10^{-4} \, \text{T} \).
- (b) The magnetic field at a point on the perpendicular bisector is approximately \( 7.7 \times 10^{-5} \, \text{T} \).