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 $B\pi r{R}_1^2={\mu }_0{N}_1\pi R_1^2$ The total flux through all the turn in a length $l$ and B is $\varphi =({\mu }_0{N}_{1}i\pi R_1^2)N_{2}l$ $=({\mu }_0{N}_1{N}_2\pi R_1^{2}l)i$ ?(i) But we know that, Comparing Eqs. (i) and (ii), we get $\varphi =Mi$ ...(ii) $M={\mu }_0{N}_1{N}_2\pi R_1^{2}l$