Relation between magnetic moment and angular velocity is -

Options

A. `Mprop omega`
B. `M prop omega^(2)`
C. `Mprop sqrt(omega)`
D. None of these

Answer

To derive the relation between magnetic moment (M) and angular velocity (ω), we can follow these steps:

### Step 1: Understand the Magnetic Moment
The magnetic moment (M) for a coil carrying current is defined as:
\[ M = I \times A \]
where \( I \) is the current flowing through the coil and \( A \) is the area of the coil.

### Step 2: Express the Area of the Coil
For a circular coil with radius \( r \), the area \( A \) can be expressed as:
\[ A = \pi r^2 \]
Thus, the magnetic moment can be rewritten as:
\[ M = I \times \pi r^2 \]

### Step 3: Relate Current to Charge and Angular Velocity
The average current \( I \) can be expressed in terms of charge \( Q \) and the time period \( T \):
\[ I = \frac{Q}{T} \]
For an electron rotating in a circular path, the angular velocity \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
From this, we can express \( T \) as:
\[ T = \frac{2\pi}{\omega} \]
Substituting this into the equation for current gives:
\[ I = \frac{Q}{\frac{2\pi}{\omega}} = \frac{Q \omega}{2\pi} \]

### Step 4: Substitute Current into Magnetic Moment Equation
Now, substitute the expression for current \( I \) back into the magnetic moment equation:
\[ M = \left(\frac{Q \omega}{2\pi}\right) \times \pi r^2 \]
This simplifies to:
\[ M = \frac{Q \omega r^2}{2} \]

### Step 5: Establish the Relation
From the equation \( M = \frac{Q \omega r^2}{2} \), we can see that the magnetic moment \( M \) is directly proportional to the angular velocity \( \omega \):
\[ M \propto \omega \]

### Conclusion
Thus, the relation between magnetic moment and angular velocity is:
\[ M \text{ is proportional to } \omega \]