Question

Find the mutual inductance between the straight wire and the square loop of figure.

Answer 1

Answer:

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Answer 2

The mutual inductance between the straight wire and the square loop is given by $M=\frac{\mu_{0} a}{2 \pi x} \ln \left[1+\frac{a}{b}\right]$

Explanation:

Step 1:

The flux or movement through the square frame is ϕ = Mi

Let's calculate the flux via the square frame first.

Now let's consider a loop dimension of length dx at a distance x from the wire.

Area of loop element, A = adx

Magnetic field x away from wire,  

$B=\frac{\mu_{0} i}{2 \pi x}$

The element's magnetic flux is indicated in

$d \phi=\frac{\mu_{0} i \times a d x}{2 \pi x}$

Step 2:

The cumulative flux through the frame is determined by

$\phi=\int d \phi$

$=\int_{b}^{a+b} \frac{\mu_{0} i \times a d x}{2 \pi x}$

$=\frac{\mu_{0} i a}{2 \pi x} \ln \left[1+\frac{a}{b}\right]$

Also,  

ϕ=Mi

Step 3:

The mutual inductivity is thus calculated as

$\begin{aligned}&M i=\frac{\mu_{0} i a}{2 \pi x} \ln \left[1+\frac{a}{b}\right]\\&M=\frac{\mu_{0} a}{2 \pi x} \ln \left[1+\frac{a}{b}\right]\end{aligned}$