Apply Lenz's law: the induced current always produces a magnetic field that opposes the change in magnetic flux through the circuit.
(a) The induced current is clockwise when the coil is viewed from the approaching magnet side; the near face of the coil becomes a south pole and opposes the approaching south pole. (b) In both coils the induced current is clockwise when each coil is viewed from the magnet between them: the left coil opposes the approaching south pole and the right coil opposes the receding north pole. (c) When the key is just closed, the flux through the right loop suddenly increases; the induced current in the right loop is in the sense that produces magnetic field opposite to the field due to the left loop. (d) The induced current in the left loop opposes the change caused by the rheostat: if the current in the right loop increases, the induced current is opposite in sense to the right-loop current as viewed along the common axis; if it decreases, the induced current is in the same sense. (e) When the key is just released, the current and flux due to the left solenoid decrease; the induced current in the right solenoid is in the sense that supports the original flux through it. (f) No induced current flows in the ring, because the magnetic field due to the straight wire is tangential to the ring and the magnetic flux through the ring is zero even while the current decreases.
In Fig. 6.16(a), the magnetic field is into the page and the loop area increases as the irregular wire becomes circular. Flux into the page increases, so the induced current must produce a field out of the page; this requires anticlockwise current. In Fig. 6.16(b), the magnetic field is out of the page and the loop area decreases as it is deformed into a narrow wire. The induced current must maintain outward flux, so it is also anticlockwise.
(a) Anticlockwise. (b) Anticlockwise.
The magnetic field in a long solenoid is $B=\mu_0nI$. Thus the induced emf in the small loop is $\epsilon=A\mu_0n\,dI/dt$. Here $n=15\,\text{cm}^{-1}=1500\,\text{m}^{-1}$, $A=2.0\times10^{-4}\,\text{m}^2$ and $dI/dt=(4.0-2.0)/0.1=20\,\text{A s}^{-1}$. Therefore $\epsilon=(2.0\times10^{-4})(4\pi\times10^{-7})(1500)(20)=7.5\times10^{-6}\,\text{V}$.
$7.5\times10^{-6}\,\text{V}$.
As the loop leaves the field, $\epsilon=B l v$, where $l$ is the side cutting the boundary. If motion is normal to the longer side, $l=8\,\text{cm}=0.08\,\text{m}$, so $\epsilon=0.3(0.08)(0.01)=2.4\times10^{-4}\,\text{V}$. The voltage lasts while the loop crosses through its shorter width, $0.02/0.01=2\,\text{s}$. If motion is normal to the shorter side, $l=0.02\,\text{m}$, so $\epsilon=0.3(0.02)(0.01)=6.0\times10^{-5}\,\text{V}$, lasting $0.08/0.01=8\,\text{s}$.
(a) $2.4\times10^{-4}\,\text{V}$ for $2\,\text{s}$. (b) $6.0\times10^{-5}\,\text{V}$ for $8\,\text{s}$.
For a rod rotating about one end in a uniform magnetic field parallel to the axis, the motional emf is $\epsilon=\frac12B\omega l^2$. With $B=0.5\,\text{T}$, $\omega=400\,\text{rad s}^{-1}$ and $l=1.0\,\text{m}$, $\epsilon=\frac12(0.5)(400)(1.0)^2=100\,\text{V}$.
$100\,\text{V}$.
The motional emf is $\epsilon=Blv=(0.30\times10^{-4})(10)(5.0)=1.5\times10^{-3}\,\text{V}$. Taking east as $+x$, north as $+y$, and upward as $+z$, the falling wire has velocity $-\hat{z}$ and the horizontal earth field is along north. For positive charges, $\mathbf{v}\times\mathbf{B}=(-\hat{z})\times\hat{y}=+\hat{x}$, so positive charge is driven towards the east end. Hence the emf is from west to east and the east end is at higher potential.
(a) $1.5\times10^{-3}\,\text{V}$. (b) The emf is directed from west to east. (c) The eastern end is at higher potential.
For self-induction, $|\epsilon|=L|dI/dt|$. The current changes by $5.0\,\text{A}$ in $0.1\,\text{s}$, so $|dI/dt|=50\,\text{A s}^{-1}$. Thus $L=200/50=4.0\,\text{H}$.
$4.0\,\text{H}$.
Flux linkage with the other coil is $N\Phi=MI$. Therefore the change in flux linkage is $\Delta(N\Phi)=M\Delta I=1.5(20-0)=30\,\text{Wb-turn}$. The time interval is not needed for flux linkage; it would be needed for induced emf.
$30\,\text{Wb-turn}$.