The magnitude of magentic field , due to a dipole of magnetic moment `2.4 Am^(2)` , at a point 200 cm away from it in the direction making an equal of `90^(@)` with the dipole axis is

Options

A. `3 xx 10^(-6) T`
B. `3 xx 10 ^(-7) T`
C. `3 xx 10^(-8) T`
D. `0.3 xx 10^(-7) T`

Answer

To find the magnitude of the magnetic field \( B \) due to a magnetic dipole at a point making an angle of \( 90^\circ \) with the dipole axis, we can use the formula for the magnetic field at the equatorial position of a magnetic dipole.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Magnetic moment \( M = 2.4 \, \text{Am}^2 \)
- Distance \( R = 200 \, \text{cm} = 2 \, \text{m} \)
- Angle \( \theta = 90^\circ \)

2. **Use the Formula for Magnetic Field at Equator**:
The magnetic field \( B \) at a point on the equatorial line of a magnetic dipole is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{M}{R^3} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Substitute the Values into the Formula**:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2.4}{(2)^3} \]
Simplifying this gives:
\[ B = 10^{-7} \cdot \frac{2.4}{8} \]

4. **Calculate the Value**:
\[ B = 10^{-7} \cdot 0.3 = 3 \times 10^{-8} \, \text{T} \]

5. **Final Answer**:
The magnitude of the magnetic field at the point is:
\[ B = 3 \times 10^{-8} \, \text{T} \]