Two concentric coils, each of radius equal to 2π cm are placed at...

The magnetic field at the centre of a circular coil is given by $B = \dfrac{\mu_0 I}{2R}$.

For the first coil, $B_1 = \dfrac{\mu_0 I_1}{2R}$.

For the second coil, $B_2 = \dfrac{\mu_0 I_2}{2R}$.

Since the coils are placed at right angles to each other, their magnetic fields at the centre are perpendicular. The resultant magnetic field is:

$B = \sqrt{B_1^2 + B_2^2} = \dfrac{\mu_0}{2R} \sqrt{I_1^2 + I_2^2}$

Substituting the given values $R = 2\pi \times 10^{-2}$ m, $I_1 = 3$ A, $I_2 = 4$ A, and $\mu_0 = 4\pi \times 10^{-7}$ T m/A:

$B = \dfrac{4\pi \times 10^{-7}}{2 \times 2\pi \times 10^{-2}} \sqrt{3^2 + 4^2}$

$B = \dfrac{4\pi \times 10^{-7}}{4\pi \times 10^{-2}} \times 5$

$B = 10^{-5} \times 5 = 5 \times 10^{-5}$ Wb/m^2$<br><br>Answer:$5 \times 10^{-5}