What is the magnitude of axial field due to a bar magnet of length 3 cm at a distance of 75 cm from its mid-point if its magnetic moment is 0.6 `Am^2`?

Options

A. 0.013`mu` T
B. 0.113 `mu` T
C. 0.213 `mu` T
D. 0.313 `mu` T

Explanation

To find the magnitude of the axial magnetic field due to a bar magnet at a certain distance, we can use the formula for the axial magnetic field \( B_A \):

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

where:
- \( B_A \) is the axial magnetic field,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( m \) is the magnetic moment of the bar magnet,
- \( r \) is the distance from the midpoint of the magnet.

### Step 1: Identify the given values
- Length of the bar magnet = 3 cm (not needed for the calculation of the field)
- Distance from the midpoint = 75 cm = 0.75 m
- Magnetic moment \( m = 0.6 \, \text{Am}^2 \)

### Step 2: Substitute the values into the formula
Convert the distance into meters and substitute the values into the formula:

\[ B_A = \frac{(4\pi \times 10^{-7}) \cdot 2(0.6)}{4\pi (0.75)^3} \]

### Step 3: Simplify the equation
Notice that \( 4\pi \) in the numerator and denominator cancels out:

\[ B_A = \frac{2 \cdot 0.6 \times 10^{-7}}{(0.75)^3} \]

### Step 4: Calculate \( (0.75)^3 \)
Calculate \( (0.75)^3 \):

\[ (0.75)^3 = 0.421875 \]

### Step 5: Substitute back and calculate \( B_A \)
Now substitute back into the equation:

\[ B_A = \frac{1.2 \times 10^{-7}}{0.421875} \]

### Step 6: Perform the division
Calculating the division gives:

\[ B_A \approx 2.84 \times 10^{-7} \, \text{T} \]

### Step 7: Convert to microtesla
Convert the result to microtesla:

\[ B_A \approx 0.284 \, \mu T \]

### Final Answer
Thus, the magnitude of the axial magnetic field at a distance of 75 cm from the midpoint of the bar magnet is approximately:

\[ B_A \approx 0.284 \, \mu T \]