Chapter 6: Electromagnetic Induction (EMI and AC)

The figures are in the NCERT book. In every case use Lenz's law: the induced current sets up a field that opposes the change in flux.

• A magnet's N pole approaching a coil: the facing end of the coil becomes N (it repels). A receding N pole: the facing end becomes S (it attracts). The reverse holds for S poles.
• A current switched on or increasing in one coil: the induced current in the neighbouring coil flows in the opposite sense. A current switched off or decreasing: the induced current flows in the same sense.

Applying these rules to the six figures:

• (a) along qrpq
• (b) along prqp in one coil and yzxy in the other
• (c) along yzxy
• (d) along zyxz
• (e) along xryx
• (f) no current: the field lines lie in the plane of the loop, so the flux through it does not change.

(a) qrpq; (b) prqp and yzxy; (c) yzxy; (d) zyxz; (e) xryx; (f) no induced current.

(a) For a given length of wire, a circle encloses the largest area. As the irregular loop becomes a circle, its area and so the flux (into the page) increase. The induced current opposes this by making a field out of the page, so it flows anticlockwise, along adcb.

(b) As the circle is squeezed into a narrow wire, its area and the flux (out of the page) decrease. The induced current tries to keep the flux by making a field out of the page, so it flows anticlockwise, along a′d′c′b′.

(a) Anticlockwise, adcb (opposing the increase in flux). (b) Anticlockwise, a′d′c′b′ (opposing the decrease in flux).

$n=15$ per cm $=1500$ m⁻¹, $A=2.0\times 10^{-4}$ m², $\frac{dI}{dt}=\frac{4.0-2.0}{0.1}=20$ A/s.

Flux through the loop: $\Phi =BA={\mu }_{0}nIA$, so

$$\epsilon ={\mu }_{0}nA\frac{dI}{dt}=4\pi \times 10^{-7}\times 1500\times 2.0\times 10^{-4}\times 20\approx 7.5\times 10^{-6}\text{V}$$

About $7.5\times 10^{-6}$ V.

$\epsilon =Blv$, where $l$ is the side that is cutting across the field boundary, and $v=0.01$ m/s.

(a) Moving normal to the longer side ($l=0.08$ m):

$\epsilon =0.3\times 0.08\times 0.01=2.4\times 10^{-4}$ V. It lasts while the 2 cm width leaves the field: $t=\frac{0.02}{0.01}=2$ s.

(b) Moving normal to the shorter side ($l=0.02$ m):

$\epsilon =0.3\times 0.02\times 0.01=0.6\times 10^{-4}$ V. It lasts while the 8 cm length leaves the field: $t=\frac{0.08}{0.01}=8$ s.

(a) $2.4\times 10^{-4}$ V for 2 s. (b) $0.6\times 10^{-4}$ V for 8 s.

Each small piece $dr$ of the rod at distance $r$ moves with speed $\omega r$, giving $d\epsilon =B\omega rdr$. Adding up from 0 to $l$:

$$\epsilon =\frac{1}{2}B\omega l^2=\frac{1}{2}\times 0.5\times 400\times (1.0{)}^2=100\text{V}$$

100 V.

(a) $\epsilon =Blv=0.30\times 10^{-4}\times 10\times 5.0=1.5\times 10^{-3}$ V.

(b) Take east $=+x$, north $=+y$, up $=+z$. The wire moves down ($\vec{v}$ along $-\hat{z}$) and ${\vec{B}}_H$ points north ($+\hat{y}$). The force on positive charges is along $\vec{v}\times \vec{B}=(-\hat{z})\times \hat{y}=+\hat{x}$, so the emf acts from west to east along the wire.

(c) Positive charges are pushed to the eastern end, so the eastern end is at the higher potential.

(a) 1.5 mV. (b) From west to east. (c) The eastern end.

$\left|\frac{dI}{dt}\right|=\frac{5.0}{0.1}=50$ A/s and $\epsilon =L\frac{dI}{dt}$, so

$$L=\frac{\epsilon }{dI/dt}=\frac{200}{50}=4\text{H}$$

L = 4 H.

The flux linkage with the second coil is $N\Phi =MI$, so

$$\Delta (N\Phi )=M\Delta I=1.5\times (20-0)=30\text{Wb}$$

(The time of 0.5 s is not needed for the change in linkage; it would give the induced emf, $\epsilon =\frac{30}{0.5}=60$ V.)

30 Wb.