Question

Two circular coils of radii $5cm$ and $10cm$ carry equal currents of $2A$. The coils have $50$ and $100$ turns respectively, and are placed in such a way that their planes as well as their centers coincide. Magnitude of magnetic field at the common center of coils is

• A. $8\pi \times {10}^{-6}$ $T$ if current in the coils are in the same sense
• B. $4\pi \times {10}^{-6}$ $T$ if current in the coils are in the opposite sense
• C. zero if current in the coils are in opposite sense
• D. $8\pi \times {10}^{-6}$ $T$ if current in the coils are in the opposite sense

Answer 1

Answer: (A), (C)

$8\pi \times {10}^{-6}$ $T$ if current in the coils are in the same sense
zero if current in the coils are in opposite sense

$B_1=n_1\frac{m{u}_0{i}_1}{2r_1}$

$B_2=n_2\frac{m{u}_0{i}_2}{2r_2}$

If currents are in the same sense:

$B=B_1+B_2=\frac{{\mu }_{0}i}{2}(\frac{n_1}{r_1}+\frac{n_2}{r_2})=\frac{4\pi \times {10}^{-7}\times 2}{2}\times (\frac{5}{50}+\frac{10}{100})\times \frac{1}{{10}^{-2}}=8\pi \times {10}^{-6}T$

If currents are in the opposite sense

$B=B_1+B_2=\frac{{\mu }_{0}i}{2}(\frac{n_1}{r_1}-\frac{n_2}{r_2})=\frac{4\pi \times {10}^{-7}\times 2}{2}\times (\frac{5}{50}-\frac{10}{100})\times \frac{1}{{10}^{-2}}=0$