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 center of a circular coil of radius $r_2$ and $N_2$ turns carrying current $I_2$ is $B_2 = \frac{\mu_0 N_2 I_2}{2 r_2}$.

• Since radius of smaller coil $r_1 \ll r_2$, magnetic field $B_2$ is nearly uniform across the entire area $A_1 = \pi r_1^2$ of the smaller coil.

• Total magnetic flux linked with smaller coil containing $N_1$ turns is $\Phi_1 = N_1 B_2 A_1$.

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

Step 2: Calculate magnetic field produced by larger coil
\[ B_2 = \frac{\mu_0 N_2 I_2}{2 r_2} \]
Substitute $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$:
\[ B_2 = \frac{(4\pi \times 10^{-7}) \times 50 \times 3}{2 \times (5 \times 10^{-2})} \]
\[ B_2 = \frac{600 \pi \times 10^{-7}}{10^{-1}} = 6\pi \times 10^{-5}\text{ T} \]

Step 3: Calculate magnetic flux linked with smaller coil
Area of smaller coil $A_1 = \pi r_1^2 = \pi (5 \times 10^{-3})^2 = 25\pi \times 10^{-6}\text{ m}^2$.
Total magnetic flux $\Phi_1$:
\[ \Phi_1 = N_1 B_2 A_1 \]
\[ \Phi_1 = 10 \times (6\pi \times 10^{-5}) \times (25\pi \times 10^{-6}) \]
\[ \Phi_1 = 1500 \pi^2 \times 10^{-11}\text{ Wb} \]
Using $\pi^2 \approx 9.87$:
\[ \Phi_1 = 1.5 \pi^2 \times 10^{-7}\text{ Wb} \approx 1.5 \times 9.87 \times 10^{-7}\text{ Wb} \approx 1.48 \times 10^{-6}\text{ Wb} \]

Step 4: Conclusion
The magnetic flux linked with the smaller coil is $1.5 \pi^2 \times 10^{-7}\text{ Wb} \approx 1.48 \times 10^{-6}\text{ Wb}$.