Two concentric coils of radius $2π\, cm$ each are placed with their planes at right angles to each other and carry currents of 3 A and 4 A, respectively. The net magnetic field in $Wb/m^2$ at the center of the coils will be

$5 × 10^{-5}$

The correct answer is Option (4) → $5 × 10^{-5}$

Given:

Radius of each coil = $r = 2\pi \text{ cm} = 2\pi \times 10^{-2} \text{ m}$

Currents in coils = $I_1 = 3 \text{ A}$, $I_2 = 4 \text{ A}$

Magnetic field at the center of a circular coil is given by:

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

For the two perpendicular coils:

$B_1 = \frac{4\pi \times 10^{-7} \times 3}{2 \times 2\pi \times 10^{-2}} = 3 \times 10^{-5} \text{ Wb/m}^2$

$B_2 = \frac{4\pi \times 10^{-7} \times 4}{2 \times 2\pi \times 10^{-2}} = 4 \times 10^{-5} \text{ Wb/m}^2$

Since the fields are perpendicular, the resultant field is:

$B = \sqrt{B_1^2 + B_2^2}$

$B = \sqrt{(3 \times 10^{-5})^2 + (4 \times 10^{-5})^2}$

$B = 5 \times 10^{-5} \text{ Wb/m}^2$

The net magnetic field at the center is $5 \times 10^{-5} \text{ Wb/m}^2$.