The magnetic moment of a short bar magnet is `2 Am ^(2).` The magnetic field at a point 1 m away from it in a direction making an angle `60^(@)` with its magnetic moment is

Options

A. `sqrt5 xx 10 ^(-7)T`
B. `10^(-7)T`
C. `sqrt7 xx 10 ^(-7)T`
D. `7 xx 10 ^(-7)T`

Answer

To solve the problem, we need to calculate the magnetic field at a point 1 meter away from a short bar magnet with a magnetic moment of \(2 \, \text{Am}^2\), making an angle of \(60^\circ\) with the magnetic moment. We will use the formula for the magnetic field \(B\) at a point due to a short bar magnet.

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic moment, \(M = 2 \, \text{Am}^2\)
- Distance from the magnet, \(r = 1 \, \text{m}\)
- Angle with the magnetic moment, \(\theta = 60^\circ\)

2. **Use the formula for the magnetic field \(B\) at a distance \(r\) from a short bar magnet:**
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3} \cdot \sqrt{1 + 3 \cos^2 \theta} \]
where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) (the permeability of free space).

3. **Substitute the known values into the formula:**
- First, calculate \(r^3\):
\[ r^3 = 1^3 = 1 \, \text{m}^3 \]
- Now substitute \(M\), \(r\), and \(\theta\) into the formula:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2}{1} \cdot \sqrt{1 + 3 \cos^2 60^\circ} \]

4. **Calculate \(\cos 60^\circ\):**
\[ \cos 60^\circ = \frac{1}{2} \]
Therefore,
\[ \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

5. **Substitute \(\cos^2 60^\circ\) into the equation:**
\[ B = 2 \cdot \sqrt{1 + 3 \cdot \frac{1}{4}} = 2 \cdot \sqrt{1 + \frac{3}{4}} = 2 \cdot \sqrt{\frac{7}{4}} \]

6. **Simplify the expression:**
\[ B = 2 \cdot \frac{\sqrt{7}}{2} = \sqrt{7} \]

7. **Final calculation of the magnetic field:**
\[ B = \sqrt{7} \times 10^{-7} \, \text{T} \]

### Conclusion:
The magnetic field at a point 1 meter away from the bar magnet, making an angle of \(60^\circ\) with its magnetic moment, is:
\[ B = \sqrt{7} \times 10^{-7} \, \text{T} \]