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

$\frac{μ_{0} a}{2 π} \ln (1 + \frac{a}{2 b})$

$\frac{μ_{0} a}{π} \ln (1 + \frac{b}{2 a})$

$\frac{μ_{0} a}{2 π} \ln (1 + \frac{a}{b})$

$\frac{μ_{0} a}{2 π} \ln (1 + \frac{b}{a})$

Solution

$\frac{μ_{0} a}{2 π} \ln (1 + \frac{a}{b})$

Magnetic field due to wire

B = $\frac{μ_{0}}{2 πx} . i$

dA = a dx

and d $ϕ$ = B. dA

On integrating both sides, we get

$\int dϕ = \int B dA$

⇒ $ϕ = \frac{μ_{0} i}{2 π}$ $\int_{b}^{a + b} \frac{a dx}{x}$

= $\frac{μ_{0} i}{2 π} a {[\ln x]}_{b}^{a + b}$

= $\frac{μ_{0} a}{2 π}$ $a [\ln (a + b) - \ln b]$

= $\frac{μ_{0}}{2 π} \ln (1 + \frac{a}{b})$

= M i

Where M = coefficient of mutual inductance

and M = $\frac{μ_{0} i}{2 π} \ln (1 + \frac{a}{b})$

For Daily Free Study Material Join wiredfaculty