NCERT Solutions: Class 12 Physics Chapter 6 - Electromagnetic Induction

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 $\epsilon=(2.0\times10^{-4})(4\pi\times10^{-7})(1500)(20)=7.5\times10^{-6}\,\text{V}$.

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 $0.08/0.01=8\,\text{s}$.

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 $l=1.0\,\text{m}$, $\epsilon=\frac12(0.5)(400)(1.0)^2=100\,\text{V}$.

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.

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 $L=200/50=4.0\,\text{H}$.

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.