Question

Two solenoids A and B are coaxially place as shown in figure below.
The radius of inner and outer solenoids are $ {{R}_{1}} $ and $ {{R}_{2}} $ respectively and the numbers of turns per unit length are $ {{N}_{1}} $ and $ {{N}_{2}} $ respectively. Consider a length $ l $ of each solenoids, calculate the mutual inductance between them.

• A. $ M={{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi R_{1}^{2}l $
• B. $ M={{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi /{{R}_{1}}{{R}_{2}} $
• C. $ M={{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi R_{1}^{2}R_{2}^{2}{{l}^{2}} $
• D. $ M={{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi \frac{R_{1}^{2}}{R_{2}^{2}}l $

Answer 1

Answer: (A)

Suppose, a current i is passed through the inner solenoid. A magnetic field $ B={{\mu }_{0}}{{N}_{1}}i $ produced inside A, whereas the field outside it is zero. The flux through each turn of A is $ B\pi rR_{1}^{2}={{\mu }_{0}}{{N}_{1}}\pi R_{1}^{2} $ The total flux through all the turn in a length $ l $ and B is $ \phi =({{\mu }_{0}}{{N}_{1}}i\pi R_{1}^{2}){{N}_{2}}l $ $ =({{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi R_{1}^{2}l)i $ ?(i) But we know that, Comparing Eqs. (i) and (ii), we get $ \phi =Mi $ ...(ii) $ M={{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi R_{1}^{2}l $