A magnet is parallel to a uniform magnetic field. If it is rotated by `60^(@)`, the work done is 0.8 J. How much work is done in moving it `30^(@)` further

Options

A. `0.8xx10^(7) ergs`
B. `0.4J`
C. `8J`
D. `0.8 ergs`

Explanation

To solve the problem, we need to understand the relationship between the work done on a magnetic dipole in a magnetic field and the angles involved.

### Step-by-Step Solution:

1. **Understanding the Work Done**:
The work done \( W \) in rotating a magnetic dipole in a magnetic field is given by the formula:
\[ W = -\mu B (\cos \theta_2 - \cos \theta_1) \]
where \( \mu \) is the magnetic moment, \( B \) is the magnetic field strength, \( \theta_1 \) is the initial angle, and \( \theta_2 \) is the final angle.

2. **Initial Rotation**:
In the first part of the problem, the magnet is rotated from \( 0^\circ \) to \( 60^\circ \) and the work done is given as \( 0.8 \, J \). Here, we can set:
- \( \theta_1 = 0^\circ \)
- \( \theta_2 = 60^\circ \)

Plugging these values into the work done formula:
\[ 0.8 = -\mu B (\cos 60^\circ - \cos 0^\circ) \]
We know that \( \cos 60^\circ = \frac{1}{2} \) and \( \cos 0^\circ = 1 \):
\[ 0.8 = -\mu B \left( \frac{1}{2} - 1 \right) = -\mu B \left( -\frac{1}{2} \right) \]
Simplifying gives:
\[ 0.8 = \frac{\mu B}{2} \]
Thus, we find:
\[ \mu B = 1.6 \, J \]

3. **Further Rotation**:
Now, we need to find the work done when the magnet is rotated from \( 60^\circ \) to \( 90^\circ \). Here:
- \( \theta_1 = 60^\circ \)
- \( \theta_2 = 90^\circ \)

Using the same work done formula:
\[ W' = -\mu B (\cos 90^\circ - \cos 60^\circ) \]
We know that \( \cos 90^\circ = 0 \):
\[ W' = -\mu B (0 - \frac{1}{2}) = \mu B \cdot \frac{1}{2} \]
Substituting \( \mu B = 1.6 \, J \):
\[ W' = 1.6 \cdot \frac{1}{2} = 0.8 \, J \]

4. **Final Answer**:
Therefore, the work done in moving the magnet \( 30^\circ \) further (from \( 60^\circ \) to \( 90^\circ \)) is:
\[ W' = 0.8 \, J \]

### Conclusion:
The work done in moving the magnet \( 30^\circ \) further is \( 0.8 \, J \).

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