A proton is revolving on a circular path of radius 2 mm with frequency 10 Hz. Magnetic dipole moment associated with proton is

Options

A. `2 xx 10^(-24) A-m^(2)`
B. `4 xx 10^(-24) A-m^(2)`
C. `3 xx 10^(-20) A-m^(2)`
D. `6 xx 10^(-20) A- m^(2)`

Explanation

To find the magnetic dipole moment associated with a proton revolving in a circular path, we can use the formula for the magnetic dipole moment (μ) of a current loop:

\[ \mu = I \cdot A \]

where:
- \( I \) is the current,
- \( A \) is the area of the circular path.

### Step 1: Calculate the Area (A) of the Circular Path
The area \( A \) of a circle is given by the formula:

\[ A = \pi r^2 \]

where \( r \) is the radius of the circular path. Given that the radius \( r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \):

\[ A = \pi (2 \times 10^{-3})^2 = \pi (4 \times 10^{-6}) = 4\pi \times 10^{-6} \, \text{m}^2 \]

### Step 2: Calculate the Current (I)
The current \( I \) can be calculated from the frequency \( f \) of the proton's revolution. The frequency is given as \( 10 \, \text{Hz} \). The current can be defined as the charge passing through a point per unit time. For a proton, the charge \( q \) is approximately \( 1.6 \times 10^{-19} \, \text{C} \).

The current \( I \) can be calculated as:

\[ I = q \cdot f \]

Substituting the values:

\[ I = (1.6 \times 10^{-19} \, \text{C}) \cdot (10 \, \text{Hz}) = 1.6 \times 10^{-18} \, \text{A} \]

### Step 3: Calculate the Magnetic Dipole Moment (μ)
Now we can substitute the values of \( I \) and \( A \) into the formula for the magnetic dipole moment:

\[ \mu = I \cdot A \]

Substituting the values:

\[ \mu = (1.6 \times 10^{-18} \, \text{A}) \cdot (4\pi \times 10^{-6} \, \text{m}^2) \]

Calculating this gives:

\[ \mu = 6.4\pi \times 10^{-24} \, \text{A m}^2 \]

Using \( \pi \approx 3.14 \):

\[ \mu \approx 6.4 \times 3.14 \times 10^{-24} \approx 20.1 \times 10^{-24} \, \text{A m}^2 \]

Thus, the magnetic dipole moment associated with the proton is approximately:

\[ \mu \approx 2.01 \times 10^{-23} \, \text{A m}^2 \]

### Final Answer
The magnetic dipole moment associated with the proton is approximately \( 2.01 \times 10^{-23} \, \text{A m}^2 \).