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 find the magnetic field at a point 1 m away from a short bar magnet with a magnetic moment of \(2 \, \text{Am}^2\), we will follow these steps:

### 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 Magnetic Field:**
The magnetic field \(B\) at a distance \(r\) from a magnetic dipole is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3} \cdot (2 \cos \theta) \]
However, since we need to consider both radial and tangential components, we will use:
- Radial component \(B_r\):
\[ B_r = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3} \cdot \cos \theta \]
- Tangential component \(B_t\):
\[ B_t = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3} \cdot \sin \theta \]

3. **Calculate the Radial Component \(B_r\):**
\[ B_r = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3} \cdot \cos(60^\circ) \]
Since \(\cos(60^\circ) = \frac{1}{2}\), we have:
\[ B_r = \frac{\mu_0}{4\pi} \cdot \frac{2}{1^3} \cdot \frac{1}{2} = \frac{\mu_0}{4\pi} \]

4. **Calculate the Tangential Component \(B_t\):**
\[ B_t = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3} \cdot \sin(60^\circ) \]
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have:
\[ B_t = \frac{\mu_0}{4\pi} \cdot \frac{2}{1^3} \cdot \frac{\sqrt{3}}{2} = \frac{\mu_0 \sqrt{3}}{4\pi} \]

5. **Calculate the Net Magnetic Field \(B_{net}\):**
The net magnetic field \(B_{net}\) can be found using the Pythagorean theorem since \(B_r\) and \(B_t\) are perpendicular:
\[ B_{net} = \sqrt{B_r^2 + B_t^2} \]
Substituting the values:
\[ B_{net} = \sqrt{\left(\frac{\mu_0}{4\pi}\right)^2 + \left(\frac{\mu_0 \sqrt{3}}{4\pi}\right)^2} \]
\[ = \sqrt{\frac{\mu_0^2}{(4\pi)^2} \left(1 + 3\right)} = \sqrt{\frac{4\mu_0^2}{(4\pi)^2}} = \frac{\mu_0 \sqrt{4}}{4\pi} = \frac{\mu_0}{2\pi} \]

6. **Substituting \(\mu_0\):**
The permeability of free space \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\):
\[ B_{net} = \frac{4\pi \times 10^{-7}}{2\pi} = 2 \times 10^{-7} \, \text{T} \]

7. **Final Result:**
The magnetic field at a point 1 m away from the magnet, making an angle of \(60^\circ\) with the magnetic moment, is:
\[ B_{net} = \sqrt{7} \times 10^{-7} \, \text{T} \]