A: The relation between magnetic moment and angular momentum is true for every finite size body.
R: Ratio of magnetic dipole moment and angular momentum is just dependent on specific charge of the body.

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion then mark (1)
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark 2.
C. If assertion is true statement but Reason is false, then mark (3)
D. If both Assertion and Reason are false statements, then mark (4)

Answer

To solve the assertion and reason question, we will analyze both statements step by step.

### Step 1: Understanding the Assertion
The assertion states that "the relation between magnetic moment and angular momentum is true for every finite size body."

- **Magnetic Moment (M)** is defined for a current loop as \( M = I \cdot A \), where \( I \) is the current and \( A \) is the area.
- For a charged particle moving in a circular path, the magnetic moment can be expressed in terms of charge, frequency, and area.

### Step 2: Deriving the Magnetic Moment
For a charged particle:
- Current \( I \) can be expressed as \( I = \frac{E}{T} \), where \( E \) is the charge and \( T \) is the time period.
- The frequency \( \nu \) is given by \( \nu = \frac{1}{T} \), thus \( I = E \cdot \nu \).

Substituting this into the formula for magnetic moment:
\[ M = I \cdot A = E \cdot \nu \cdot A \]

For a circular path, the area \( A \) is \( \pi r^2 \):
\[ M = E \cdot \nu \cdot \pi r^2 \]

### Step 3: Relating Frequency and Angular Velocity
The frequency \( \nu \) can also be related to angular velocity \( \omega \):
\[ \nu = \frac{\omega}{2\pi} \]

### Step 4: Expressing Angular Momentum
Angular momentum \( L \) for a particle is given by:
\[ L = m \cdot v \cdot r \]
where \( v \) is the linear velocity.

### Step 5: Establishing the Ratio
Now, we can derive the ratio of magnetic moment to angular momentum:
1. Substitute \( v = R \cdot \omega \) into the equations.
2. From the earlier derived expressions, we can write:
\[ \frac{M}{L} = \frac{E \cdot \nu \cdot \pi r^2}{m \cdot v \cdot r} \]

After simplification, we find:
\[ \frac{M}{L} = \frac{E}{2m} \]
This shows that the ratio of magnetic moment to angular momentum depends only on the specific charge \( \frac{E}{m} \).

### Step 6: Conclusion
Both the assertion and reason are true. The assertion is correct as the relation holds for every finite size body, and the reason correctly explains that the ratio of magnetic moment to angular momentum is dependent on the specific charge of the body.

### Final Answer
Both the assertion (A) and reason (R) are true, and R correctly explains A.

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