In a uniform magnetic field, the magnetic needle has a magnetic moment `9.85 xx 10^(-2) A//m^(2)` and moment of inertia `5 xx 10^(-6 ) "kg m"^(2)`. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is ________ mT.
[Take `pi^(2)` as 9.85]

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Time Period of Oscillation
Given that the magnetic needle performs 10 complete oscillations in 5 seconds, we can calculate the time period \( T \) of one oscillation.

\[ T = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{5 \, \text{s}}{10} = 0.5 \, \text{s} \]

### Step 2: Use the Formula for Time Period in a Magnetic Field
The time period \( T \) of a magnetic needle in a magnetic field is given by the formula:

\[ T = 2\pi \sqrt{\frac{I}{mB}} \]

Where:
- \( I \) is the moment of inertia
- \( m \) is the magnetic moment
- \( B \) is the magnetic field strength

### Step 3: Rearranging the Formula to Solve for \( B \)
We can rearrange the formula to solve for \( B \):

\[ B = \frac{4\pi^2 I}{m T^2} \]

### Step 4: Substitute the Known Values
We know:
- \( I = 5 \times 10^{-6} \, \text{kg m}^2 \)
- \( m = 9.85 \times 10^{-2} \, \text{A m}^2 \)
- \( T = 0.5 \, \text{s} \)
- \( \pi^2 = 9.85 \)

Substituting these values into the equation gives:

\[ B = \frac{4 \times 9.85 \times 5 \times 10^{-6}}{9.85 \times (0.5)^2} \]

### Step 5: Simplifying the Expression
First, calculate \( (0.5)^2 \):

\[ (0.5)^2 = 0.25 \]

Now substitute this back into the equation:

\[ B = \frac{4 \times 9.85 \times 5 \times 10^{-6}}{9.85 \times 0.25} \]

The \( 9.85 \) cancels out:

\[ B = \frac{4 \times 5 \times 10^{-6}}{0.25} \]

Calculating the numerator:

\[ 4 \times 5 = 20 \]

Now substitute back:

\[ B = \frac{20 \times 10^{-6}}{0.25} \]

### Step 6: Calculate the Final Value
To divide by \( 0.25 \), we can multiply by \( 4 \):

\[ B = 20 \times 10^{-6} \times 4 = 80 \times 10^{-6} \, \text{T} \]

### Step 7: Convert to milliTesla
Since \( 1 \, \text{T} = 1000 \, \text{mT} \):

\[ B = 80 \times 10^{-6} \, \text{T} = 80 \, \text{mT} \]

### Final Answer
The magnitude of the magnetic field is:

\[ \boxed{8 \, \text{mT}} \]

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