A magnetic dipole of magnetic moment `6xx 10^(-2) A m^2` and moment of inertia `12 xx 10^(-6) kg m^2` performs oscillations in a magnetic field of `2 xx 10^(-2)` T. The time taken by the dipole to complete 20 oscillations is `(pi~~3)`

Options

A. 18 s
B. 6 s
C. 36 s
D. 12 s

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the formula for the time period of oscillation of a magnetic dipole
The time period \( T \) of a magnetic dipole in a magnetic field is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{\mu B}} \]
where:
- \( I \) is the moment of inertia,
- \( \mu \) is the magnetic moment,
- \( B \) is the magnetic field strength.

### Step 2: Substitute the given values into the formula
Given:
- Magnetic moment \( \mu = 6 \times 10^{-2} \, \text{A m}^2 \)
- Moment of inertia \( I = 12 \times 10^{-6} \, \text{kg m}^2 \)
- Magnetic field \( B = 2 \times 10^{-2} \, \text{T} \)

Substituting these values into the formula:
\[ T = 2\pi \sqrt{\frac{12 \times 10^{-6}}{(6 \times 10^{-2})(2 \times 10^{-2})}} \]

### Step 3: Simplify the expression inside the square root
Calculating the denominator:
\[ (6 \times 10^{-2})(2 \times 10^{-2}) = 12 \times 10^{-4} \]
Now, substituting this back into the equation:
\[ T = 2\pi \sqrt{\frac{12 \times 10^{-6}}{12 \times 10^{-4}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{10^{-6}}{10^{-4}}} = 2\pi \sqrt{10^{-2}} = 2\pi \times 10^{-1} = 0.2\pi \, \text{s} \]

### Step 4: Calculate the time for 20 oscillations
The time taken for 20 oscillations \( T' \) is given by:
\[ T' = 20 \times T = 20 \times 0.2\pi = 4\pi \, \text{s} \]
Given that \( \pi \approx 3 \), we can approximate:
\[ T' \approx 4 \times 3 = 12 \, \text{s} \]

### Final Answer
The time taken by the dipole to complete 20 oscillations is approximately:
\[ \boxed{12 \, \text{s}} \]