Two concentric coils each of radius equal to `2 pi cm` are placed at right angles to each other ` 3 ampere and 4 ampere` are the currents flowing in each coil respectively. The magnetic induction in `weber//m^(2)` at the centre of the coils will be
`( mu_(0) = 4 pi xx 10^(-7) Wb// A.m)`

Options

A. `5xx10^(-5)`
B. `12xx10^(-5)`
C. `7xx10^(-5)`
D. `10^(-5)`

Answer

To find the magnetic induction at the center of two concentric coils placed at right angles to each other, we can follow these steps:

### Step 1: Identify the Given Values
- Radius of each coil, \( R = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Current in the first coil, \( I_1 = 3 \, \text{A} \)
- Current in the second coil, \( I_2 = 4 \, \text{A} \)
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{Wb/A.m} \)

### Step 2: Calculate the Magnetic Field Due to Each Coil
The magnetic field \( B \) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]

#### For the first coil (with current \( I_1 \)):
\[ B_1 = \frac{\mu_0 I_1}{2R} = \frac{4\pi \times 10^{-7} \times 3}{2 \times 0.02} \]
Calculating this:
\[ B_1 = \frac{4\pi \times 10^{-7} \times 3}{0.04} = \frac{12\pi \times 10^{-7}}{0.04} = 3\pi \times 10^{-5} \, \text{Wb/m}^2 \]

#### For the second coil (with current \( I_2 \)):
\[ B_2 = \frac{\mu_0 I_2}{2R} = \frac{4\pi \times 10^{-7} \times 4}{2 \times 0.02} \]
Calculating this:
\[ B_2 = \frac{4\pi \times 10^{-7} \times 4}{0.04} = \frac{16\pi \times 10^{-7}}{0.04} = 4\pi \times 10^{-5} \, \text{Wb/m}^2 \]

### Step 3: Calculate the Resultant Magnetic Field
Since the coils are placed at right angles to each other, the resultant magnetic field \( B_{\text{net}} \) can be found using the Pythagorean theorem:
\[ B_{\text{net}} = \sqrt{B_1^2 + B_2^2} \]

Substituting the values:
\[ B_{\text{net}} = \sqrt{(3\pi \times 10^{-5})^2 + (4\pi \times 10^{-5})^2} \]
Calculating this:
\[ B_{\text{net}} = \sqrt{(9\pi^2 \times 10^{-10}) + (16\pi^2 \times 10^{-10})} = \sqrt{25\pi^2 \times 10^{-10}} = 5\pi \times 10^{-5} \, \text{Wb/m}^2 \]

### Step 4: Final Result
Using \( \pi \approx 3.14 \):
\[ B_{\text{net}} \approx 5 \times 3.14 \times 10^{-5} \approx 1.57 \times 10^{-4} \, \text{Wb/m}^2 \]

Thus, the magnetic induction at the center of the coils is approximately:
\[ B_{\text{net}} \approx 5 \times 10^{-5} \, \text{Wb/m}^2 \]