The magnitude of the equatorial magnetic field due to a bar magnet of length 2 cm at a distance of 1m from its mid-point is (magnetic moment of the bar magnet is 0.60 A m)

Options

A. `5.0xx10^(-5)` T
B. `6.0xx10^(-8)` T
C. `7.0xx10^(-7)` T
D. `8.0xx10^(-8)` T

Explanation

To solve the problem of finding the magnitude of the equatorial magnetic field due to a bar magnet, we will follow these steps:

### Step 1: Identify the Given Values
- Length of the bar magnet, \( L = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Distance from the midpoint of the magnet, \( D = 1 \, \text{m} \)
- Magnetic moment of the bar magnet, \( M = 0.60 \, \text{A m} \)

### Step 2: Determine the Formula for the Equatorial Magnetic Field
The formula for the equatorial magnetic field \( B_E \) due to a bar magnet is given by:
\[ B_E = \frac{\mu_0}{4\pi} \cdot \frac{M}{D^3} \]
where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 3: Substitute the Values into the Formula
Since the length \( L \) of the bar magnet is small compared to the distance \( D \), we can neglect \( L \) in the formula. Thus, we will use:
\[ B_E = \frac{\mu_0}{4\pi} \cdot \frac{M}{D^3} \]

Substituting the values:
\[ B_E = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{0.60}{(1)^3} \]

### Step 4: Simplify the Expression
The \( 4\pi \) in the numerator and denominator cancels out:
\[ B_E = 10^{-7} \cdot 0.60 \]

### Step 5: Calculate the Result
Now, calculate the value:
\[ B_E = 0.60 \times 10^{-7} = 6 \times 10^{-8} \, \text{T} \]

### Final Answer
The magnitude of the equatorial magnetic field due to the bar magnet at a distance of 1 m from its midpoint is:
\[ B_E = 6 \times 10^{-8} \, \text{T} \]

---