Question

Two circular coils of radii 5.0 cm and 10 cm carry equal currents of 2.0 A. The coils have 50 and 100 turns respectively and are placed in such a way that their planes as well as the centres coincide. Find the magnitude of the magnetic field B at the common centre of the coils if the currents in the coils are (a) in the same sense (b) in the opposite sense.

Answer 1

No. of turns: n₁ = 50 and n₂ = 100
Magnitude of currents: i₁ = i₂ = 2 A
Radii of loops: r₁ = 5 cm and r₂ = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre is given by
$$\begin{gathered} B=\frac{{\mathrm{\mu }}_0{n}_1{i}_1}{2r_1}+\frac{{\mathrm{\mu }}_0{n}_2{i}_2}{2r_2} \\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 50\times 2}{2\times 5\times {10}^{-2}}+\frac{4\mathrm{\pi }\times {10}^{-7}\times 100\times 2}{2\times 10\times {10}^{-2}} \\ =4\mathrm{\pi }\times {10}^{-4}+4\mathrm{\pi }\times {10}^{-4} \\ =2\times 4\mathrm{\pi }\times {10}^{-6} \\ =8\mathrm{\pi }\times {10}^{-3}\mathrm{T} \end{gathered}$$

(b) In the opposite sense:
The magnetic field intensity at the centre is given by

$$\begin{gathered} B=\frac{{\mathrm{\mu }}_0{n}_1{i}_1}{2r_1}-\frac{{\mathrm{\mu }}_0{n}_2{i}_2}{2r_2} \\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 50\times 2}{2\times 5\times {10}^{-2}}-\frac{4\mathrm{\pi }\times {10}^{-7}\times 100\times 2}{2\times 10\times {10}^{-2}} \\ =0 \end{gathered}$$