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

\[B = \frac{\mu_0 n_1 i_1}{2 r_1} + \frac{\mu_0 n_2 i_2}{2 r_2}\]
\[ = \frac{4\pi \times {10}^{- 7} \times 50 \times 2}{2 \times 5 \times {10}^{- 2}} + \frac{4\pi \times {10}^{- 7} \times 100 \times 2}{2 \times 10 \times {10}^{- 2}}\]
\[ = 4\pi \times {10}^{- 4} + 4\pi \times {10}^{- 4} \]
\[ = 2 \times 4\pi \times {10}^{- 6} \]
\[ = 8\pi \times {10}^{- 3} T\]

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

\[B = \frac{\mu_0 n_1 i_1}{2 r_1} - \frac{\mu_0 n_2 i_2}{2 r_2}\]
\[ = \frac{4\pi \times {10}^{- 7} \times 50 \times 2}{2 \times 5 \times {10}^{- 2}} - \frac{4\pi \times {10}^{- 7} \times 100 \times 2}{2 \times 10 \times {10}^{- 2}} \]
\[ = 0\]