Two concentric coils X and Y, each of radius 2 cm, are placed at right angles to each other. If X and Y carry 4 A and 3 A currents, respectively, then the value of the magnetic field (in \( \mathrm{Wb\,m^{-2}} \)) at the centre of the coils is (Take \( \mu_0 = 4\pi \times 10^{-7}\,\mathrm{Wb\,A^{-1}\,m^{-1}} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 5\pi \times 10^{-5} \)
The field at the centre of a circular coil is \( B = \mu_0 I/(2r) \) (Biot-Savart law; each coil is taken as a single turn, since no number of turns is given). The field is along the axis of the coil. The coils are at right angles, so their fields are perpendicular and add as vectors.
\( \displaystyle r = 2\,\mathrm{cm} = 0.02\,\mathrm{m} \)
\( \displaystyle B_X = \frac{\mu_0 I_X}{2r} = \frac{4\pi\times10^{-7} \times 4}{2 \times 0.02} = 4\pi\times10^{-5}\,\mathrm{Wb\,m^{-2}} \)
\( \displaystyle B_Y = \frac{\mu_0 I_Y}{2r} = \frac{4\pi\times10^{-7} \times 3}{2 \times 0.02} = 3\pi\times10^{-5}\,\mathrm{Wb\,m^{-2}} \)
\( \displaystyle B = \sqrt{B_X^{2} + B_Y^{2}} = \sqrt{(4\pi)^{2} + (3\pi)^{2}}\times10^{-5} = 5\pi\times10^{-5}\,\mathrm{Wb\,m^{-2}} \)