91Ó°ÊÓ

The mutual inductance of two solenoids is given by $\mathbf{M}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}{\mathbf{AN}}_{\mathbf{1}}{\mathbf{N}}_{\mathbf{2}}}{\mathbf{l}}$where A is the cross-sectional area of the inner solenoid and I is the length of the outer solenoid.

Two solenoids, the first one has a length of I =50.0 cm and contains N= 6750 coils, and the second one is 3.0 cm long and D =0.120 cm in diameter and contains N=15 coils. The second one is centered inside the first one.Substitute the values in the above equation we have,

$$\begin{gathered} \mathrm{M}=\frac{\left(4\mathrm{π}×{10}^{-7}\mathrm{Tm}/\mathrm{A}\right)\mathrm{π}\left(6.00×{10}^{14}\mathrm{m}\right)×6750×15}{0.500\mathrm{m}} \\ =2.88×{10}^{-7}\mathrm{H} \end{gathered}$$

Therefore,the mutual inductance of solenoid is$2.88×{10}^{-7}\mathrm{H}$

If the current in the outer solenoid is changing at di/dt=49.2 A/s. Then the induced emf in the inner solenoid is,

$$\begin{gathered} {\mathrm{Î\mu }}_2=\mathrm{M}\left|\frac{{\mathrm{di}}_1}{\mathrm{dt}}\right| \\ =2.88×{10}^{-7}\mathrm{H}×\left(49.2\mathrm{A}/\mathrm{s}\right) \\ =1.42×{10}^{-5}\mathrm{V} \end{gathered}$$

Therefore,the emf induced in the inner solenoid is$1.42×{10}^{-5}\mathrm{V}$