Answer
See answers.
Work Step by Step
Use equation 21–2b to find the emf induced in the short coil. Here N is 14, the number of loops in the short coil. The relevant area is the solenoid’s cross-sectional area, and the magnetic field B is the field created by the solenoid. The current changing in the solenoid causes the flux to change.
$$|\epsilon|=N_{short}\frac{|A\Delta B|}{\Delta t}= N_{short}\frac{|A\Delta (\mu_oIN_{solenoid})/\mathcal{l}_{solenoid}|}{\Delta t}$$
$$= N_{short}\frac{ N_{solenoid}\mu_o A}{\mathcal{l}_{solenoid}}\frac{\Delta I}{\Delta t}$$
$$= 14\frac{(600)(4\pi \times10^{-7})\pi(0.0125m)^2 }{0.25m}\frac{5.0A}{0.60s}=1.7\times10^{-4}V$$
The current will flow so as to create a field that opposes the field inside the solenoid.
