A magnet of magnetic moment `20` C.G.S. units is freely suspended in a uniform magnetic field of intensity 0.3 C.G.S. units. The amount of work done in deflecting it by an angle of `30^(@)` in C.G.S. unit is

Options

A. 6
B. `3sqrt(3)`
C. 3`(2-sqrt(3))`
D. 3

Answer

To solve the problem of calculating the work done in deflecting a magnet by an angle of \(30^\circ\), we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Values:**
- Magnetic moment \(M = 20\) C.G.S. units
- Magnetic field intensity \(B = 0.3\) C.G.S. units
- Angle of deflection \(\theta = 30^\circ\)

2. **Formula for Work Done:**
The work done \(W\) in deflecting a magnetic moment in a magnetic field is given by the formula:
\[ W = MB(1 - \cos \theta) \]

3. **Calculate \(\cos \theta\):**
For \(\theta = 30^\circ\):
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]

4. **Substituting the Values into the Formula:**
Now, substituting the values into the work done formula:
\[ W = 20 \times 0.3 \times \left(1 - \frac{\sqrt{3}}{2}\right) \]

5. **Simplifying the Expression:**
First, calculate \(20 \times 0.3\):
\[ 20 \times 0.3 = 6 \]
Now substituting this back into the equation:
\[ W = 6 \times \left(1 - \frac{\sqrt{3}}{2}\right) \]

6. **Finding a Common Denominator:**
To simplify \(1 - \frac{\sqrt{3}}{2}\):
\[ 1 = \frac{2}{2} \quad \text{so,} \quad 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \]

7. **Final Calculation:**
Now substituting back:
\[ W = 6 \times \frac{2 - \sqrt{3}}{2} = \frac{6(2 - \sqrt{3})}{2} = 3(2 - \sqrt{3}) \]

8. **Conclusion:**
Therefore, the amount of work done in deflecting the magnet by an angle of \(30^\circ\) is:
\[ W = 3(2 - \sqrt{3}) \text{ C.G.S. units} \]