A) \[\frac{{{\mu }_{0}}a}{2\pi }ln\left( 1-\frac{b}{c} \right)\]
B) \[\frac{{{\mu }_{0}}a}{2\pi }ln\left( 1+\frac{b}{c} \right)\]
C) \[\frac{{{\mu }_{0}}a}{\pi }ln\left( 1+\frac{b}{c} \right)\]
D) \[\frac{{{\mu }_{0}}a}{\sqrt{2}\pi }ln\left( 1+\frac{b}{c} \right)\]
Correct Answer: B
Solution :
[b] Let current \[{{i}_{1}}\]in straight wire be upward. Then the magnetic field due to the straight wire has magnitude \[{{B}_{1}}={{\mu }_{0}}{{i}_{1}}/2\pi r\] at distance r. In accordance with right hand rule, \[{{B}_{1}}\]points inward to the plane of page. We consider a differential strip of thickness dr, area\[d{{A}_{2}}=a\,\,\,dr\]. Magnetic flux through area dA, \[d{{\phi }_{B}}={{B}_{1}}\] (a dr). Total flux through the loop, \[{{\phi }_{B}}=\int{{{B}_{1}}ad\,r=\int_{c}^{c+b}{\frac{{{\mu }_{0}}{{i}_{1}}}{2\pi r}a\,\,dr}}\] \[=\frac{{{\mu }_{0}}{{i}_{1}}a}{2\pi }\int_{a}^{c+b}{\frac{dr}{r}=\frac{{{\mu }_{0}}{{i}_{1}}a}{2\pi }\ln \left( \frac{c+b}{c} \right)}\] Therefore mutual inductance, \[M=\frac{\phi }{{{i}_{1}}}=\frac{{{\mu }_{0}}a}{2\pi }\ln \left( 1+\frac{b}{c} \right)\]
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