A circular coil of 25 turns and radius of `20cm` carrying a current of `1 A` rests with its plane normal to an external field of magnitude `5.0 xx 10^(-2)T`. The coil is free to tum about-an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of `2 s^(-1)`. The moment of inertia of the coil about its axis of rotation is `(x xx 10^(-2))/(16 pi) kg m^(2)`. Find `x`

Answer

To solve the problem, we need to find the moment of inertia \( I \) of the circular coil given the oscillation frequency and the parameters of the coil. Here’s a step-by-step solution:

### Step 1: Identify the given parameters
- Number of turns \( N = 25 \)
- Radius of the coil \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Current \( I = 1 \, \text{A} \)
- External magnetic field \( B = 5.0 \times 10^{-2} \, \text{T} \)
- Frequency of oscillation \( f = 2 \, \text{s}^{-1} \)

### Step 2: Calculate the magnetic moment \( \mu \) of the coil
The magnetic moment \( \mu \) of a circular coil is given by the formula:
\[ \mu = N \cdot I \cdot A \]
where \( A \) is the area of the coil. The area \( A \) of a circle is given by:
\[ A = \pi r^2 \]
Substituting the values:
\[ A = \pi (0.2)^2 = \pi \cdot 0.04 = 0.04\pi \, \text{m}^2 \]
Now substituting into the magnetic moment formula:
\[ \mu = 25 \cdot 1 \cdot 0.04\pi = 1.0\pi \, \text{A m}^2 \]

### Step 3: Relate the frequency of oscillation to the moment of inertia
The frequency \( f \) is related to the moment of inertia \( I \) and the magnetic field \( B \) by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{mB}{I}} \]
Rearranging this to find \( I \):
\[ I = \frac{mB}{(2\pi f)^2} \]

### Step 4: Calculate the effective mass \( m \)
The effective mass \( m \) can be calculated using the formula:
\[ m = \frac{\mu B}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). However, we can also use the fact that the torque due to the magnetic field is equal to the restoring torque due to gravity to find an equivalent mass.

But for this problem, we can directly substitute \( \mu \) and \( B \) into the formula for \( I \) without needing to find \( m \) explicitly.

### Step 5: Substitute values into the equation for \( I \)
Using the calculated values:
\[ I = \frac{(1.0\pi)(5.0 \times 10^{-2})}{(2\pi \cdot 2)^2} \]
Calculating the denominator:
\[ (2\pi \cdot 2)^2 = (4\pi)^2 = 16\pi^2 \]
Now substituting back:
\[ I = \frac{(1.0\pi)(5.0 \times 10^{-2})}{16\pi^2} = \frac{5.0 \times 10^{-2}}{16\pi} \]

### Step 6: Express \( I \) in the required form
We need to express \( I \) in the form \( \frac{x \times 10^{-2}}{16\pi} \):
\[ I = \frac{5.0}{16\pi} \times 10^{-2} \]
Thus, comparing with the required form, we find:
\[ x = 5.0 \]

### Final Answer
The value of \( x \) is \( 5 \).