Q94 / 180 · JIPMER 2016 · Physics
In the given figure what will be the coefficient of mutual inductance
- A$\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{2 b}\right)$
- B$\frac{\mu_0 a}{\pi} \ln \left(1+\frac{b}{2 a}\right)$
- Correct answerC$\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{b}\right)$
- D$\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{b}{a}\right)$
Solution
Magnetic field due to wire
$\begin{aligned} B & =\frac{\mu_0}{2 \pi x} \cdot i \\ d A & =a d x\end{aligned}$
and $\quad d \phi=B \cdot d A$
On integrating both sides,
$\begin{aligned} \int d \phi & =\int B d A \\ \phi & =\frac{\mu_0 i}{2 \pi} \int_b^{a+b} \frac{a d x}{x} \\ & =\left.\frac{\mu_0 i}{2 \pi} a \ln x\right|_b ^{a+b} \\ & =\frac{\mu_0 i}{2 \pi} a[\ln (a+b)-\ln b] \\ & =\frac{\mu_0 i}{2 \pi} a\left[\ln \left(1+\frac{a}{b}\right)\right]=M i\end{aligned}$
where, $M=$ coefficient of mutual inductance
$\therefore \quad M=\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{b}\right)$
and $\quad d \phi=B \cdot d A$
On integrating both sides,
$\begin{aligned} \int d \phi & =\int B d A \\ \phi & =\frac{\mu_0 i}{2 \pi} \int_b^{a+b} \frac{a d x}{x} \\ & =\left.\frac{\mu_0 i}{2 \pi} a \ln x\right|_b ^{a+b} \\ & =\frac{\mu_0 i}{2 \pi} a[\ln (a+b)-\ln b] \\ & =\frac{\mu_0 i}{2 \pi} a\left[\ln \left(1+\frac{a}{b}\right)\right]=M i\end{aligned}$
where, $M=$ coefficient of mutual inductance
$\therefore \quad M=\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{b}\right)$Hence, the correct answer is option: 'C'.