Q152 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Physics
A circular coil consists 70 closely wound turns and has a radius of $10 \mathrm{~cm}$. An externally produced magnetic field of magnitude $2 \times 10^{-3} \mathrm{~T}$ is applied perpendicular to the coil. The net flux through the coil is found to vanish when the current in the coil is $2.2 \mathrm{~A}$. The inductance of the coil is
- Correct answerA$2 \mathrm{mH}$
- B$3 \mathrm{mH}$
- C$4 \mathrm{mH}$
- D$1.5 \mathrm{mH}$
Solution
Given, number of turns in the circular coil,
$N=70, r=10 \mathrm{~cm}=10^{-1} \mathrm{~m}$
$B=2 \times 10^{-3} \mathrm{~T}$
$\theta=0^{\circ}$
Magnetic flux, $\phi=N B A \cos \theta$
$=70 \times 2 \times 10^{-3} \times \pi r^2 \times \cos 0^{\circ}$
$=140 \pi \times 10^{-3} \times\left(10^{-1}\right)^2$
$=140 \pi \times 10^{-5}$
$=140 \times \frac{22}{7} \times 10^{-5}=4.4 \times 10^{-3} \mathrm{~Wb}$
We know that,
$\phi=L I$
$L=\frac{\phi}{I}=\frac{4.4 \times 10^{-3}}{2.2}=2 \times 10^{-3} \mathrm{H}=2 \mathrm{mH}$
Hence, the correct answer is option: 'A'.