Search any question & find its solution
Question:
Answered & Verified by Expert
Magnetic field induction at the centre of a circular coil of radius $5 \mathrm{~cm}$ and carrying a current $0.9 \mathrm{~A}$ is (in SI units) $\left(\varepsilon_0=\right.$ absolute permittivity of air in SI units, velocity of light $=3 \times 10^8 \mathrm{~ms}^{-1}$ )
Options:
Solution:
1621 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{\varepsilon_0 10^{16}}$
Magnetic field induction at the centre of circular coil,
$\begin{aligned}
& c^2=\frac{1}{\mu_0 \varepsilon_0} \\
& \therefore \quad \mu_0=\frac{1}{\varepsilon_0 c^2} \\
&
\end{aligned}$
So, from Eq. (i), we get
$\begin{aligned}
B & =\frac{1}{\varepsilon_0 c^2} \frac{I}{2 r} \\
& =\frac{1}{\varepsilon_0 \times\left(3 \times 10^8\right)^2} \times \frac{0.9}{2 \times 5 \times 10^{-2}} \\
& =\frac{1 \times 9}{\varepsilon_0 \times 9 \times 10^{16}}=\frac{1}{\varepsilon_0 \times 10^{16}}
\end{aligned}$
$\begin{aligned}
& c^2=\frac{1}{\mu_0 \varepsilon_0} \\
& \therefore \quad \mu_0=\frac{1}{\varepsilon_0 c^2} \\
&
\end{aligned}$
So, from Eq. (i), we get
$\begin{aligned}
B & =\frac{1}{\varepsilon_0 c^2} \frac{I}{2 r} \\
& =\frac{1}{\varepsilon_0 \times\left(3 \times 10^8\right)^2} \times \frac{0.9}{2 \times 5 \times 10^{-2}} \\
& =\frac{1 \times 9}{\varepsilon_0 \times 9 \times 10^{16}}=\frac{1}{\varepsilon_0 \times 10^{16}}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.