Search any question & find its solution
Question:
Answered & Verified by Expert
A coil having 2000 turns is wound tightly in the form of a spiral with inner and outer radii $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$, respectively. When a current $\frac{1}{\pi} \mathrm{mA}$ passes through the coil, then the magnetic field at the centre is calculated be $K \ln 3 \times 10^{-6} \mathrm{~T}$. The value of $K$ is
Options:
Solution:
1970 Upvotes
Verified Answer
The correct answer is:
36
Number of turns, $N=2000$
Inner radius, $r_1=1 \mathrm{~cm}$
Outer radius, $r_2=3 \mathrm{~cm}$
Current, $I=\frac{1}{\pi} \mathrm{mA}=\frac{10^{-3}}{\pi} \mathrm{A}$
Magnetic field at the centre of the coil, $B=\frac{\mu_0 N I}{2 r}$
$\begin{aligned}\text{Mean radius,} \quad r & =\frac{r_1+r_2}{2}=\frac{1+3}{2} \\ & =2 \mathrm{~cm}=0.02 \mathrm{~m}\end{aligned}$
$$
\begin{array}{rlrl}
\therefore B =\frac{4 \pi \times 10^{-7} \times 2000 \times 10^{-3}}{2 \times 0.02 \times \pi} \\
=4 \times 10^{-5} \mathrm{~T} \\
\Rightarrow K \ln 3 \times 10^{-6}=4 \times 10^{-5} \\
\Rightarrow K =\frac{40}{\ln 3}=36.4 \simeq 36
\end{array}
$$
Inner radius, $r_1=1 \mathrm{~cm}$
Outer radius, $r_2=3 \mathrm{~cm}$
Current, $I=\frac{1}{\pi} \mathrm{mA}=\frac{10^{-3}}{\pi} \mathrm{A}$
Magnetic field at the centre of the coil, $B=\frac{\mu_0 N I}{2 r}$
$\begin{aligned}\text{Mean radius,} \quad r & =\frac{r_1+r_2}{2}=\frac{1+3}{2} \\ & =2 \mathrm{~cm}=0.02 \mathrm{~m}\end{aligned}$
$$
\begin{array}{rlrl}
\therefore B =\frac{4 \pi \times 10^{-7} \times 2000 \times 10^{-3}}{2 \times 0.02 \times \pi} \\
=4 \times 10^{-5} \mathrm{~T} \\
\Rightarrow K \ln 3 \times 10^{-6}=4 \times 10^{-5} \\
\Rightarrow K =\frac{40}{\ln 3}=36.4 \simeq 36
\end{array}
$$
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.