Question

Two coils, one of radius $0.5\text{ cm}$ having $10$ turns and the other of radius $5\text{ cm}$ having $50$ turns are placed coaxially in air such that their centres are coincident. Calculate the magnetic flux through the smaller coil if the larger coil carries a current of $3\text{ A}$.

Hint:

Always calculate field of the LARGER coil at the center first, because field across small area $A_1$ is uniform. Trying to calculate field of smaller coil over larger area is mathematically much harder!

Answer 1

Concept:
• Magnetic field at the common centre of a circular coil of radius \(r_2\) with \(N_2\) turns carrying current \(I_2\): \(B_2 = \frac{\mu_0 N_2 I_2}{2 r_2}\).

• Since \(r_1 \ll r_2\), the field of the larger coil is nearly uniform over the area \(A_1 = \pi r_1^2\) of the smaller coil.

• Flux linked with the smaller coil of \(N_1\) turns: \(\Phi_1 = N_1 B_2 A_1\).

Step 1: Write the given data
Smaller coil: \(r_1 = 0.5\text{ cm} = 5 \times 10^{-3}\text{ m}\), \(N_1 = 10\).
Larger coil: \(r_2 = 5\text{ cm} = 5 \times 10^{-2}\text{ m}\), \(N_2 = 50\), \(I_2 = 3\text{ A}\).

Step 2: Magnetic field at the centre due to the larger coil
\[ B_2 = \frac{\mu_0 N_2 I_2}{2 r_2} = \frac{(4\pi \times 10^{-7}) \times 50 \times 3}{2 \times 5 \times 10^{-2}} \]
\[ B_2 = \frac{600\pi \times 10^{-7}}{10^{-1}} = 6\pi \times 10^{-4}\text{ T} \approx 1.88 \times 10^{-3}\text{ T} \]

Step 3: Flux through the smaller coil
Area of the smaller coil: \(A_1 = \pi r_1^2 = \pi (5 \times 10^{-3})^2 = 25\pi \times 10^{-6}\text{ m}^2\).
\[ \Phi_1 = N_1 B_2 A_1 = 10 \times (6\pi \times 10^{-4}) \times (25\pi \times 10^{-6}) \]
\[ \Phi_1 = 1500\pi^2 \times 10^{-10}\text{ Wb} = 1.5\pi^2 \times 10^{-7}\text{ Wb} \]
Using \(\pi^2 \approx 9.87\):
\[ \Phi_1 \approx 1.5 \times 9.87 \times 10^{-7}\text{ Wb} \approx 1.48 \times 10^{-6}\text{ Wb} \]

Final Answer:
The magnetic flux through the smaller coil is \(\Phi_1 = 1.5\pi^2 \times 10^{-7}\text{ Wb} \approx 1.48 \times 10^{-6}\text{ Wb}\).