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A small square loop of side ' a ' and one turn is placed inside a larger square loop of side b and one turn (\mathrm{b}>>\mathrm{a}). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b' then the coefficient of mutual inductance between the two loops is :
 
Option: 1 \begin{aligned} &\frac{\mu_{0}}{4 \pi} 8 \sqrt{2} \frac{a^{2}}{b} \\ \end{aligned}
Option: 2 \frac{\mu_{0}}{4 \pi} 8 \sqrt{2} \frac{b^{2}}{a} \\
Option: 3 \frac{\mu_{0}}{4 \pi} \frac{8 \sqrt{2}}{b} \\
Option: 4 \frac{\mu_{0}}{4 \pi} \frac{8 \sqrt{2}}{a}

Answers (1)

best_answer


Magnetic field due to square loop at its cenre 0
B_{0}= 4\left [ \frac{\mu _{0}I}{4\pi \left ( \frac{b}{2} \right )}\times \left ( \sin 45^{\circ} +\sin 45^{\circ}\right ) \right ]
B_{0}= \frac{2\mu _{0}I}{\pi b}\times \sqrt{2}= \frac{2\sqrt{2}\mu _{0}I}{\pi b}
\phi = B_{0}A= B_{0}a^{2}= \frac{2\sqrt{2}\mu _{0}Ia^{2}}{\pi b}
\phi = MI=\frac{8\sqrt{2}\mu _{0}Ia^{2}}{4\pi b}
\therefore M= \frac{8\sqrt{2}\mu _{0}a^{2}}{4\pi b}

Posted by

vishal kumar

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