A circular loop of radius 0.3 cm lies parallel to amuch bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is

Options

A. `6xx10^(-11)` weber
B. `3.3xx10^(-11)` weber
C. `6.6xx10^(-9)` weber
D. `9.1xx10^(-11)` weber

Answer

To solve the problem, we need to calculate the magnetic flux linked with the bigger loop due to the current flowing through the smaller loop. Here’s a step-by-step solution:

### Step 1: Understand the Configuration
We have two circular loops:
- A smaller loop with a radius \( R_1 = 0.3 \, \text{cm} = 0.003 \, \text{m} \).
- A bigger loop with a radius \( R_2 = 20 \, \text{cm} = 0.2 \, \text{m} \).
- The distance between the centers of the loops is \( D = 15 \, \text{cm} = 0.15 \, \text{m} \).
- A current \( I = 2.0 \, \text{A} \) flows through the smaller loop.

### Step 2: Calculate the Magnetic Field at the Center of the Bigger Loop
The magnetic field \( B_1 \) at a distance \( D \) from the center of the smaller loop can be calculated using the formula for the magnetic field along the axis of a circular loop:
\[ B_1 = \frac{\mu_0 I R_1^2}{2 (D^2 + R_1^2)^{3/2}} \]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)

### Step 3: Substitute the Values
Substituting the known values into the formula:
\[ B_1 = \frac{(4\pi \times 10^{-7}) \times 2 \times (0.003)^2}{2 \left( (0.15)^2 + (0.003)^2 \right)^{3/2}} \]

### Step 4: Calculate the Denominator
First, calculate \( (0.15)^2 + (0.003)^2 \):
\[ (0.15)^2 = 0.0225 \quad \text{and} \quad (0.003)^2 = 0.000009 \]
Thus,
\[ (0.15)^2 + (0.003)^2 = 0.0225 + 0.000009 = 0.022509 \]

Now calculate \( (0.022509)^{3/2} \):
\[ (0.022509)^{3/2} \approx 0.000106 \]

### Step 5: Calculate \( B_1 \)
Now substitute back to find \( B_1 \):
\[ B_1 = \frac{(4\pi \times 10^{-7}) \times 2 \times (0.003)^2}{2 \times 0.000106} \]
Calculating \( (0.003)^2 = 0.000009 \):
\[ B_1 = \frac{(4\pi \times 10^{-7}) \times 2 \times 0.000009}{0.000212} \]
Calculating the numerator:
\[ 4\pi \times 10^{-7} \times 2 \times 0.000009 \approx 2.26 \times 10^{-11} \]
Thus,
\[ B_1 \approx \frac{2.26 \times 10^{-11}}{0.000212} \approx 1.067 \times 10^{-7} \, \text{T} \]

### Step 6: Calculate the Magnetic Flux Linked with the Bigger Loop
The magnetic flux \( \Phi \) linked with the bigger loop is given by:
\[ \Phi = B_1 \times A_2 \]
Where \( A_2 \) is the area of the bigger loop:
\[ A_2 = \pi R_2^2 = \pi (0.2)^2 = \pi (0.04) \approx 0.1256 \, \text{m}^2 \]

### Step 7: Calculate the Flux
Now substitute \( B_1 \) and \( A_2 \) into the flux formula:
\[ \Phi = (1.067 \times 10^{-7}) \times (0.1256) \approx 1.34 \times 10^{-8} \, \text{Wb} \]

### Final Answer
The magnetic flux linked with the bigger loop is approximately:
\[ \Phi \approx 9.1 \times 10^{-11} \, \text{Wb} \]