Two concentric coils X and Y, each of radius 2 cm, are placed at right angles to each other. If A and B carry 4 A and 3 A currents, respectively, then the value of the magnetic field (in $Wb\, m^{-2}$) at the centre of the coils is (Take $μ_0=4π × 10^{-7}\, Wb\, A^{-1}m^{-1}$)

$5 × 10^{-5}$

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

Magnetic field at the centre of a circular coil:

$B = \frac{\mu_{0} \, I}{2R}$

For coil X:

$I_{X} = 4 \, A, \; R = 0.02 \, m$

$B_{X} = \frac{4 \pi \times 10^{-7} \times 4}{2 \times 0.02}$

$B_{X} = \frac{16 \pi \times 10^{-7}}{0.04} = 4 \pi \times 10^{-5} \, Wb \, m^{-2}$

For coil Y:

$I_{Y} = 3 \, A, \; R = 0.02 \, m$

$B_{Y} = \frac{4 \pi \times 10^{-7} \times 3}{2 \times 0.02}$

$B_{Y} = \frac{12 \pi \times 10^{-7}}{0.04} = 3 \pi \times 10^{-5} \, Wb \, m^{-2}$

Resultant field: (since the fields are perpendicular)

$B = \sqrt{B_{X}^{2} + B_{Y}^{2}}$

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

$B = \pi \times 10^{-5} \sqrt{16 + 9}$

$B = \pi \times 10^{-5} \sqrt{25}$

$B = 5 \pi \times 10^{-5} \, Wb \, m^{-2}$

Answer: $5 \pi \times 10^{-5} \, Wb \, m^{-2}$