Two circular loops of diameters $0.6 \mathrm{~cm}$ and $40 \mathrm{~cm}$ are kept coaxially with a…
Two circular loops of diameters $0.6 \mathrm{~cm}$ and $40 \mathrm{~cm}$ are kept coaxially with a separation of $15 \mathrm{~cm}$ between their centres. If a current $2 \mathrm{~A}$ flows through the smaller loop, then the flux linked with the bigger loop is (approximately)
$9 \times 10^{-11} \mathrm{~Wb}$
$0.9 \times 10^{-11} \mathrm{~Wb}$
$1.8 \times 10^{-11} \mathrm{~Wb}$
$0.42 \times 10^{-11} \mathrm{~Wb}$
Solution
(None of the option is matching)
Magnetic field intensity due to small loop at location of larger loop is
$
B_1=\frac{\mu_0 I r^2}{2 x^3}=\frac{4 \pi \times 10^{-7} \times 2 \times\left(0.3 \times 10^{-2}\right)^2}{2 \times\left(15 \times 10^{-2}\right)^3}=\frac{36 \pi}{15^3} \times 10^{-9} \mathrm{~T}
$
Flux linked with larger loop is
$
\begin{aligned}
\phi_2=B_1 A_2= & \frac{86 \pi}{15^3} \times 10^{-19} \times \pi \times\left(20 \times 10^{-2}\right)^2 \\
& =\frac{36 \pi^2}{15^3} \times 4 \times 10^{-11}=0.42 \times 10^{-11} \mathrm{~Wb}
\end{aligned}
$