Electromagnetic Induction — NCERT Solutions

Since the figures cannot be seen directly, the standard NCERT descriptions and answers for each part are given below, based on the well-known content of these figures.

Concept used: Lenz's Law — the induced current opposes the change in magnetic flux through the loop.

(a) A coil is moved towards a bar magnet (North pole facing the coil).

• The flux through the coil increases as it moves towards the magnet.
• To oppose the increase, the induced current must create a magnetic field opposing the magnet's field (i.e., facing the magnet with a North pole).
• Direction: The induced current flows in the coil in the anti-clockwise direction when viewed from the magnet side (i.e., along qrpq in the coil).

(b) A coil is moved away from a bar magnet (South pole facing the coil).

• The flux through the coil decreases as it moves away.
• To oppose the decrease, the induced current must attract the magnet, so it creates a South pole facing the magnet.
• Direction: The induced current flows clockwise when viewed from the magnet side (i.e., along prqp).

(c) A wire loop is placed near a solenoid carrying increasing current.

• The magnetic flux through the loop due to the solenoid increases.
• By Lenz's law, the induced current opposes the increase.
• Direction: The induced current in the loop flows in the anti-clockwise direction (as viewed from the solenoid end), i.e., along yzxy.

(d) A wire loop is placed near a solenoid carrying decreasing current.

• The magnetic flux through the loop decreases.
• By Lenz's law, the induced current opposes the decrease.
• Direction: The induced current flows clockwise (as viewed from the solenoid end), i.e., along zyxz (opposite to case (c)).

(e) A rectangular loop is moved into a region of uniform magnetic field directed into the page.

• As the loop enters the field region, the flux through it increases (into the page).
• By Lenz's law, the induced current must create flux out of the page inside the loop.
• Direction: The induced current flows anti-clockwise, i.e., along adcba.

(f) Two circular loops — current in the outer loop is increased.

• The increasing current in the outer loop increases the flux through the inner loop.
• By Lenz's law, the induced current in the inner loop opposes this increase.
• Direction: The induced current in the inner loop flows in the clockwise direction (opposite to the current in the outer loop).

Concept used: Lenz's Law — the induced current opposes the change in magnetic flux. A circular loop encloses the maximum area for a given perimeter, while a straight wire encloses zero area.

(a) Wire of irregular shape turning into a circular shape:

Given: A wire loop of irregular shape is deformed into a circular shape in a uniform magnetic field directed into the page (inward).

• A circle encloses the maximum area for a given perimeter.
• As the wire turns circular, the area enclosed increases, so the magnetic flux through the loop increases (into the page).
• By Lenz's law, the induced current must oppose this increase, i.e., it must create a magnetic field out of the page inside the loop.
• Using the right-hand rule, the current must flow anti-clockwise.

$$\boxed{\text{Induced current flows anti-clockwise (a d c b a).}}$$

(b) Circular loop being deformed into a narrow straight wire:

Given: A circular loop is deformed into a narrow straight wire in a uniform magnetic field directed into the page.

• As the loop is deformed into a straight wire, the area enclosed decreases towards zero, so the magnetic flux decreases.
• By Lenz's law, the induced current must oppose this decrease, i.e., it must create a magnetic field into the page inside the loop.
• Using the right-hand rule, the current must flow clockwise.

$$\boxed{\text{Induced current flows clockwise (a d c b a).}}$$

Given:

• Number of turns per unit length of solenoid: $n = 15\,\text{turns/cm} = 15 \times 10^2\,\text{turns/m} = 1500\,\text{turns/m}$
• Area of small loop: $A = 2.0\,\text{cm}^2 = 2.0 \times 10^{-4}\,\text{m}^2$
• Change in current: $dI = 4.0 - 2.0 = 2.0\,\text{A}$
• Time interval: $dt = 0.1\,\text{s}$

Concept/Formula used:

The magnetic field inside a solenoid is:
$$B = \mu_0 n I$$

The flux through the small loop placed inside the solenoid (normal to axis):
$$\Phi = B \cdot A = \mu_0 n I A$$

By Faraday's law, the induced emf:
$$\varepsilon = -\frac{d\Phi}{dt} = -\mu_0 n A \frac{dI}{dt}$$

Calculation:
$$\frac{dI}{dt} = \frac{2.0}{0.1} = 20\,\text{A/s}$$

$$|\varepsilon| = \mu_0 \times n \times A \times \frac{dI}{dt}$$

$$|\varepsilon| = 4\pi \times 10^{-7} \times 1500 \times 2.0 \times 10^{-4} \times 20$$

$$|\varepsilon| = 4\pi \times 10^{-7} \times 1500 \times 4.0 \times 10^{-3}$$

$$|\varepsilon| = 4\pi \times 10^{-7} \times 6.0$$

$$|\varepsilon| = 4 \times 3.14159 \times 6.0 \times 10^{-7}$$

$$|\varepsilon| = 75.4 \times 10^{-7}\,\text{V}$$

$$\boxed{|\varepsilon| \approx 7.54 \times 10^{-6}\,\text{V}}$$

The induced emf in the loop is approximately $7.54 \times 10^{-6}\,\text{V}$ (or $7.54\,\mu\text{V}$).

Given:

• Sides of rectangular loop: $l_1 = 8\,\text{cm} = 0.08\,\text{m}$, $l_2 = 2\,\text{cm} = 0.02\,\text{m}$
• Magnetic field: $B = 0.3\,\text{T}$ (normal to loop)
• Velocity: $v = 1\,\text{cm s}^{-1} = 0.01\,\text{m s}^{-1}$

Concept/Formula used:
Motional emf: $\varepsilon = Blv$, where $l$ is the length of the side cutting through the field boundary.

(a) Velocity normal to the longer side (8 cm side):

The side of the loop cutting through the field boundary is the longer side ($l_1 = 0.08\,\text{m}$).

$$\varepsilon = B l_1 v = 0.3 \times 0.08 \times 0.01$$

$$\boxed{\varepsilon = 2.4 \times 10^{-4}\,\text{V}}$$

Duration: The loop must travel a distance equal to the shorter side ($l_2 = 0.02\,\text{m}$) to completely exit the field.

$$t = \frac{l_2}{v} = \frac{0.02}{0.01} = 2\,\text{s}$$

$$\boxed{t = 2\,\text{s}}$$

(b) Velocity normal to the shorter side (2 cm side):

The side of the loop cutting through the field boundary is the shorter side ($l_2 = 0.02\,\text{m}$).

$$\varepsilon = B l_2 v = 0.3 \times 0.02 \times 0.01$$

$$\boxed{\varepsilon = 6 \times 10^{-5}\,\text{V}}$$

Duration: The loop must travel a distance equal to the longer side ($l_1 = 0.08\,\text{m}$) to completely exit the field.

$$t = \frac{l_1}{v} = \frac{0.08}{0.01} = 8\,\text{s}$$

$$\boxed{t = 8\,\text{s}}$$