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) \[M={{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi \frac{R_{1}^{2}}{R_{2}^{2}}l\]
Correct Answer: A
Solution :
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) \[M={{\mu }_{0}}{{N}_{1}}{{N}_{2}}\pi R_{1}^{2}l\]You need to login to perform this action.
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