Two circular coils `A` and `B` of radius `(5)/sqrt2cm` amd `5cm` respectively current `5amp` and `(5)/sqrt2Amp` respectively The plane of `B` is perpendicular to plane of `A` their centres coincide Find the magnetic field at the centre .

Options

A. 0
B. `4pisqrt2xx10^-5T`
C. `4pixx10^-5T`
D. `2pisqrt2xx10^-5T`

Explanation

To find the magnetic field at the center of two circular coils A and B, we will follow these steps:

### Step 1: Identify the given values
- Radius of coil A, \( r_1 = \frac{5}{\sqrt{2}} \) cm = \( \frac{5}{\sqrt{2}} \times 10^{-2} \) m
- Radius of coil B, \( r_2 = 5 \) cm = \( 5 \times 10^{-2} \) m
- Current in coil A, \( I_1 = 5 \) A
- Current in coil B, \( I_2 = \frac{5}{\sqrt{2}} \) A
- The angle between the planes of the coils, \( \theta = 90^\circ \)

### Step 2: Calculate the magnetic field due to coil A at the center
The formula for the magnetic field at the center of a circular coil is given by:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

For coil A:
\[ B_1 = \frac{4\pi \times 10^{-7} \times 5}{2 \times \left(\frac{5}{\sqrt{2}} \times 10^{-2}\right)} \]

Calculating \( B_1 \):
\[ B_1 = \frac{4\pi \times 10^{-7} \times 5}{2 \times \frac{5}{\sqrt{2}} \times 10^{-2}} = \frac{4\pi \times 10^{-7} \times 5 \sqrt{2}}{10 \times 10^{-2}} = 2\sqrt{2} \pi \times 10^{-5} \, \text{T} \]

### Step 3: Calculate the magnetic field due to coil B at the center
For coil B:
\[ B_2 = \frac{4\pi \times 10^{-7} \times \frac{5}{\sqrt{2}}}{2 \times (5 \times 10^{-2})} \]

Calculating \( B_2 \):
\[ B_2 = \frac{4\pi \times 10^{-7} \times \frac{5}{\sqrt{2}}}{10 \times 10^{-2}} = \frac{4\pi \times 10^{-7} \times 5}{10\sqrt{2} \times 10^{-2}} = 2\sqrt{2} \pi \times 10^{-5} \, \text{T} \]

### Step 4: Determine the resultant magnetic field
Since the magnetic fields \( B_1 \) and \( B_2 \) are perpendicular to each other, we can use the Pythagorean theorem to find the resultant magnetic field \( B \):
\[ B = \sqrt{B_1^2 + B_2^2} \]
Substituting the values:
\[ B = \sqrt{(2\sqrt{2} \pi \times 10^{-5})^2 + (2\sqrt{2} \pi \times 10^{-5})^2} \]
\[ B = \sqrt{2 \times (2\sqrt{2} \pi \times 10^{-5})^2} = 2\sqrt{2} \pi \times 10^{-5} \sqrt{2} \]
\[ B = 4\pi \times 10^{-5} \, \text{T} \]

### Final Answer
The resultant magnetic field at the center is:
\[ B = 4\pi \times 10^{-5} \, \text{T} \]