To find the net magnetic induction at the center of the two coils, we can start by calculating the magnetic field produced by each coil separately.

For a coil with radius r and N turns carrying a current I, the magnetic field at the center of the coil is given by:

$$ B = \mu_0 \frac{NI}{2r} $$

Given that the first coil A carries a current I and the second coil B carries a current $$2I$$, the magnetic fields due to each coil at the center can be calculated as:

For coil A:

$$ B_A = \mu_0 \frac{NI}{2r} $$

For coil B:

$$ B_B = \mu_0 \frac{N(2I)}{2r} = \mu_0 \frac{2NI}{2r} = \mu_0 \frac{NI}{r} $$

Since the coils are placed concentrically with their planes perpendicular to each other, the magnetic fields at the center will be perpendicular to each other. Therefore, the net magnetic induction at the center can be found by taking the vector sum of the two magnetic fields.

The net magnetic induction B can thus be calculated using the Pythagorean theorem:

$$ B = \sqrt{B_A^2 + B_B^2} $$

Substituting the values of $$B_A$$ and $$B_B$$, we get:

$$ B = \sqrt{\left(\mu_0 \frac{NI}{2r}\right)^2 + \left(\mu_0 \frac{NI}{r}\right)^2} $$

Simplifying inside the square root:

$$ B = \sqrt{\left(\mu_0 \frac{NI}{2r}\right)^2 + \left(\mu_0 \frac{NI}{r}\right)^2} $$

$$ B = \sqrt{\left(\mu_0 \frac{NI}{2r}\right)^2 + \left(2\cdot\mu_0 \frac{NI}{2r}\right)^2} $$

$$ B = \sqrt{\left(\mu_0 \frac{NI}{2r}\right)^2 + \left(\mu_0 \frac{2NI}{2r}\right)^2} $$

$$ B = \sqrt{\left(\mu_0 \frac{NI}{2r}\right)^2 + 4\left(\mu_0 \frac{NI}{2r}\right)^2} $$

$$ B = \sqrt{1\left(\mu_0 \frac{NI}{2r}\right)^2 + 4\left(\mu_0 \frac{NI}{2r}\right)^2} $$

$$ B = \sqrt{5\left(\mu_0 \frac{NI}{2r}\right)^2} $$

$$ B = \sqrt{5}\left(\mu_0 \frac{NI}{2r}\right) $$

Hence, the correct answer is:

$$ \text{Option B: } \sqrt{5}\left(\mu_0 \frac{NI}{2r}\right) $$

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