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

Explanation

To solve the problem, we need to calculate the magnetic induction at the center of two concentric coils placed at right angles to each other, with given currents flowing through them.

### Step-by-Step Solution:

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

2. **Calculate the Magnetic Field Due to Coil 1:**
The magnetic field \( B_1 \) at the center of a circular coil is given by the formula:
\[ B_1 = \frac{\mu_0 I_1}{2r} \]
Substituting the values:
\[ B_1 = \frac{4 \pi \times 10^{-7} \times 3}{2 \times (2 \pi \times 10^{-2})} \]
Simplifying:
\[ B_1 = \frac{4 \pi \times 10^{-7} \times 3}{4 \pi \times 10^{-2}} = \frac{3 \times 10^{-7}}{10^{-2}} = 3 \times 10^{-5} \, \text{Wb/m}^2 \]

3. **Calculate the Magnetic Field Due to Coil 2:**
Similarly, for coil 2:
\[ B_2 = \frac{\mu_0 I_2}{2r} \]
Substituting the values:
\[ B_2 = \frac{4 \pi \times 10^{-7} \times 4}{2 \times (2 \pi \times 10^{-2})} \]
Simplifying:
\[ B_2 = \frac{4 \pi \times 10^{-7} \times 4}{4 \pi \times 10^{-2}} = \frac{4 \times 10^{-7}}{10^{-2}} = 4 \times 10^{-5} \, \text{Wb/m}^2 \]

4. **Determine the Resultant Magnetic Field:**
Since the two coils are at right angles to each other, the resultant magnetic field \( B \) can be calculated using the Pythagorean theorem:
\[ B = \sqrt{B_1^2 + B_2^2} \]
Substituting the values:
\[ B = \sqrt{(3 \times 10^{-5})^2 + (4 \times 10^{-5})^2} \]
Calculating:
\[ B = \sqrt{9 \times 10^{-10} + 16 \times 10^{-10}} = \sqrt{25 \times 10^{-10}} = 5 \times 10^{-5} \, \text{Wb/m}^2 \]

5. **Final Answer:**
The magnetic induction at the center of the coils is:
\[ B = 5 \times 10^{-5} \, \text{Wb/m}^2 \]