Question

One solenoid is cantered inside another. The outer onehas a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter andcontains 15 coils. The current in the outer solenoid is changing at49.2 A/s. (a) What is the mutual inductance of these solenoids?(b) Find the emf induced in the inner solenoid.

Short answer

A)The mutual inductance of solenoid is $2.88\times {10}^{-7}\mathrm{H}$

B) The emf induced in the inner solenoid is$1.42\times {10}^{-5}\mathrm{V}$

Step-by-step solution

Concept of the mutual inductance of solenoid

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

Calculate the mutual inductance of 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,

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

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

Calculate the emf induced in the inner solenoid.

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,

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

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