Electromagnetic Induction — NCERT Solutions

The chapter gives only the exercise prompt, not the detailed figure information needed to apply Lenz’s law case by case.

From Lenz’s law, the induced current opposes the change in magnetic flux. - (a) Irregular wire turning into a circular shape: the loop area increases while it becomes circular, so the induced current is such that it opposes the increase in flux. - (b) Circular loop deformed into a narrow straight wire: the loop area decreases, so the induced current is such that it opposes the decrease in flux. So only the opposing nature can be stated from the given chapter text.

For a long solenoid, magnetic field inside is
$$B=\mu_0 n I$$
So the flux through the small loop is
$$\Phi = BA = \mu_0 n I A$$
Hence induced emf is
$$\varepsilon = A\mu_0 n \frac{dI}{dt}$$
Given:

• turns per cm = 15, so $n=15\times 100=1500\,\text{m}^{-1}$
• area $A=2.0\,\text{cm}^2=2.0\times10^{-4}\,\text{m}^2$
• current change $\Delta I=4.0-2.0=2.0\,\text{A}$
• time $\Delta t=0.1\,\text{s}$

So
$$\frac{dI}{dt}=\frac{2.0}{0.1}=20\,\text{A s}^{-1}$$
Now
$$\varepsilon = (2.0\times10^{-4})(4\pi\times10^{-7})(1500)(20)$$
$$=2.0\times10^{-4}\times 4\pi\times10^{-7}\times 3.0\times10^{4}$$
$$=2.0\times10^{-4}\times 3.77\times10^{-2} \approx 7.5\times10^{-6}\,\text{V}$$
So the induced emf is about $7.5\,\mu\text{V}$.

Use motional emf:
$$\varepsilon = Blv$$
Here

• $B=0.3\,\text{T}$
• velocity $v=1\,\text{cm s}^{-1}=0.01\,\text{m s}^{-1}$

The length $l$ is the side perpendicular to the motion.

(a) Motion normal to the longer side

Then the side cutting the field lines is the longer side, $l=8\,\text{cm}=0.08\,\text{m}$.
$$\varepsilon = 0.3\times0.08\times0.01=2.4\times10^{-4}\,\text{V}$$
Time for the loop to leave the field region equals width in direction of motion divided by speed. If it moves normal to the longer side, it crosses the shorter side width $2\,\text{cm}=0.02\,\text{m}$:
$$t=\frac{0.02}{0.01}=2\,\text{s}$$

(b) Motion normal to the shorter side

Then the effective length is the shorter side, $l=2\,\text{cm}=0.02\,\text{m}$.
$$\varepsilon = 0.3\times0.02\times0.01=6.0\times10^{-5}\,\text{V}$$
Now the loop crosses the longer side width $8\,\text{cm}=0.08\,\text{m}$:
$$t=\frac{0.08}{0.01}=8\,\text{s}$$
So the induced voltage is $2.4\times10^{-4}\,\text{V}$ for 2 s in (a), and $6.0\times10^{-5}\,\text{V}$ for 8 s in (b).