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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Electromagnetic Induction

JIPMER 2016 · Physics · Electromagnetic Induction

In the given figure what will be the coefficient of mutual...

In the given figure what will be the coefficient of mutual inductance In the given figure what will be the coefficient of mutual inductance

Options

  1. Aμ0a2πln⁡(1+a2b)\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{2 b}\right)
  2. Bμ0aπln⁡(1+b2a)\frac{\mu_0 a}{\pi} \ln \left(1+\frac{b}{2 a}\right)
  3. Cμ0a2πln⁡(1+ab)\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{b}\right)
  4. Dμ0a2πln⁡(1+ba)\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{b}{a}\right)

Correct answer

C. μ0a2πln⁡(1+ab)\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{b}\right)

Step-by-step solution

Magnetic field due to wire B=μ02πx⋅idA=adx\begin{aligned} B & =\frac{\mu_0}{2 \pi x} \cdot i \\ d A & =a d x\end{aligned} In the given figure what will be the coefficient of mutual inductance and dϕ=B⋅dA\quad d \phi=B \cdot d A On integrating both sides, ∫dϕ=∫BdAϕ=μ0i2π∫ba+badxx=μ0i2πaln⁡x∣ba+b=μ0i2πa[ln⁡(a+b)−ln⁡b]=μ0i2πa[ln⁡(1+ab)]=Mi\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=M= coefficient of mutual inductance ∴M=μ0a2πln⁡(1+ab)\therefore \quad M=\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{b}\right)
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