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  • Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.

Q6.16 (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.

            

Answers (1)

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Here let's take a small element dy in the loop at y distance from the wire

Area of this element dy :

dA=a*dy

The magnetic field at dy (which is y distance away from the wire)

B=\frac{\mu _0I}{2\pi y}

The magnetic field associated with this element dy

d\phi =BdA

d\phi=\frac{\mu _0I}{2\pi y}*ady=\frac{\mu _0Ia}{2\pi}\frac{dy}{y}

\phi=\int_{x}^{a+x}\frac{\mu _0Ia}{2\pi}\frac{dy}{y}=\frac{\mu _0Ia}{2\pi}[lnx]^{a+x}_a=\frac{\mu _0Ia}{2\pi}ln[\frac{a+x}{x}]

 

Now As we know

\phi=MI where M is the mutual inductance 

so

\phi=MI=\frac{\mu _0Ia}{2\pi}ln[\frac{a+x}{x}]

M=\frac{\mu _0a}{2\pi}ln[\frac{a+x}{x}]

Hence mutual inductance between the wire and the loop is:

 \frac{\mu _0a}{2\pi}ln[\frac{a+x}{x}]

 

Posted by

Pankaj Sanodiya

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