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. `0.8 xx10^(7)`erg
B. 0.4J
C. 8J
D. 8reg

Answer

To solve the problem step by step, we will use the formula for work done in rotating a magnetic moment in a magnetic field:

**Step 1: Understand the formula for work done**
The work done (W) in rotating a magnetic moment (M) in a magnetic field (B) from an angle θ1 to θ2 is given by the formula:
\[ W = -MB (\cos \theta_2 - \cos \theta_1) \]
where:
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field strength,
- \( \theta_1 \) is the initial angle,
- \( \theta_2 \) is the final angle.

**Step 2: Identify the initial conditions**
In the first case, the magnet is initially parallel to the magnetic field. Thus:
- \( \theta_1 = 0^\circ \)
- \( \theta_2 = 60^\circ \)
- The work done \( W_1 = 0.8 \, \text{J} \)

**Step 3: Substitute values into the formula for the first case**
Using the formula:
\[ 0.8 = -MB (\cos 60^\circ - \cos 0^\circ) \]
We know:
- \( \cos 0^\circ = 1 \)
- \( \cos 60^\circ = \frac{1}{2} \)

Substituting these values:
\[ 0.8 = -MB \left( \frac{1}{2} - 1 \right) \]
\[ 0.8 = -MB \left( -\frac{1}{2} \right) \]
\[ 0.8 = \frac{MB}{2} \]

**Step 4: Solve for MB**
Multiplying both sides by 2:
\[ MB = 1.6 \, \text{N m} \]

**Step 5: Calculate work done for the second case**
Now, we need to find the work done when the magnet is rotated from \( 60^\circ \) to \( 90^\circ \):
- New \( \theta_1 = 60^\circ \)
- New \( \theta_2 = 90^\circ \)

Using the formula again:
\[ W_2 = -MB (\cos 90^\circ - \cos 60^\circ) \]
We know:
- \( \cos 90^\circ = 0 \)
- \( \cos 60^\circ = \frac{1}{2} \)

Substituting these values:
\[ W_2 = -MB (0 - \frac{1}{2}) \]
\[ W_2 = MB \cdot \frac{1}{2} \]

**Step 6: Substitute the value of MB**
Now, substituting \( MB = 1.6 \, \text{N m} \):
\[ W_2 = 1.6 \cdot \frac{1}{2} \]
\[ W_2 = 0.8 \, \text{J} \]

**Final Answer:**
The work done in moving the magnet by \( 30^\circ \) further is \( 0.8 \, \text{J} \).

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