question_answer

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

Answer:

As the square loop is moving away from the straight wire, there is a change in magnetic flux linked between the wire and the loop, hence an emf is induced in the loop and this phenomenon is called mutual induction. The flux linked with the loop µcurrent in the wire \[\phi \propto i\]                       ??? (i) \[\phi =Mi\]                                ??? (ii) Where, M is a constant and is called mutual induction. Consider a rectangular strip of small width \[dp\]at a distance p from the wire. Area of the strip \[=dp\times a\] Magnetic field intensity on the strip is \[dB=\frac{{{\mu }_{0}}}{2\pi }\frac{I}{p}\] Magnetic field linked with the strip \[d\phi =dB.\text{Area}\] \[=\frac{{{\mu }_{0}}I}{2\pi p}\times dp\times a\] The total magnetic flux linked with the square loop is given by \[\phi =\int{d\phi }=\int\limits_{x}^{a+x}{\frac{{{\mu }_{0}}Ia}{2\pi }\frac{dp}{p}}\] \[\Rightarrow \]   \[\phi =\frac{{{\mu }_{0}}Ia}{2\pi }\,\,[In\,\,p]_{x}^{a+x}=\frac{{{\mu }_{0}}Ia}{2\pi }\,\,[In\,\,(a+x)-In\,\,x]\] i.e.        \[\phi =\frac{{{\mu }_{0}}Ia}{2\pi }\,\,In\,\,\left( \frac{a+x}{x} \right)=\frac{{{\mu }_{0}}Ia}{2\pi }\,\,In\,\,\left( 1+\frac{a}{x} \right)\] Comparing Eqs. (i) and (ii), we get Mutual inductance, \[M=\frac{{{\mu }_{0}}a}{2\pi }\,\,In\,\,\left( 1+\frac{a}{x} \right)\]