Electromagnetic Induction

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.

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(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.

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

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

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(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.

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

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

Given:
- Length of rod: $l = 1.0\,\text{m}$
- Angular frequency: $\omega = 400\,\text{rad s}^{-1}$
- Magnetic field: $B = 0.5\,\text{T}$ (parallel to the axis of rotation)

Concept/Formula used:

When a rod of length $l$ rotates about one end with angular frequency $\omega$ in a uniform magnetic field $B$ (parallel to the axis), the emf developed between the centre (pivot) and the free end is:

$$\varepsilon = \frac{1}{2} B \omega l^2$$

Derivation (brief): Consider a small element $dx$ at distance $x$ from the pivot. Its velocity is $v = \omega x$. The emf across this element is $d\varepsilon = Bv\,dx = B\omega x\,dx$. Integrating from $0$ to $l$:

$$\varepsilon = \int_0^l B\omega x\,dx = B\omega \frac{l^2}{2} = \frac{1}{2}B\omega l^2$$

Calculation:

$$\varepsilon = \frac{1}{2} \times 0.5 \times 400 \times (1.0)^2$$

$$\varepsilon = \frac{1}{2} \times 0.5 \times 400$$

$$\varepsilon = \frac{200}{2} = 100\,\text{V}$$

$$\boxed{\varepsilon = 100\,\text{V}}$$

Given:
- Length of wire: $l = 10\,\text{m}$ (extending east to west)
- Speed of falling: $v = 5.0\,\text{m s}^{-1}$ (downward, i.e., vertically downward)
- Horizontal component of Earth's magnetic field: $B_H = 0.30 \times 10^{-4}\,\text{T}$ (directed from south to north, i.e., towards geographic north)

Concept/Formula used:
Motional emf: $\varepsilon = B_H l v$

The wire falls vertically downward, perpendicular to the horizontal component of Earth's field (which points north). The wire lies east–west. So the velocity ($v$, downward), the field ($B_H$, northward), and the wire (east–west) are mutually perpendicular — ideal for motional emf.

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(a) Instantaneous emf:

$$\varepsilon = B_H \times l \times v$$

$$\varepsilon = 0.30 \times 10^{-4} \times 10 \times 5.0$$

$$\varepsilon = 0.30 \times 10^{-4} \times 50$$

$$\boxed{\varepsilon = 1.5 \times 10^{-3}\,\text{V} = 1.5\,\text{mV}}$$

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(b) Direction of the emf:

Using the relation $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$:
- $\mathbf{v}$ is directed downward (south in vertical sense, but here vertically downward)
- $\mathbf{B}_H$ is directed towards the North (horizontal)
- $\mathbf{v} \times \mathbf{B}_H$: downward $\times$ north = west direction (by right-hand rule: $\hat{(-z)} \times \hat{y} = -\hat{x}$, i.e., west if x is east)

For positive charges in the wire, the force is directed towards the west end. So conventional current flows from east to west inside the wire, meaning the emf drives current from west to east in the external circuit.

$$\boxed{\text{The direction of induced emf is from west to east (i.e., the west end is at higher potential).}}$$

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(c) Higher potential end:

Positive charges accumulate at the west end of the wire (since the magnetic force on positive charges is towards west).

$$\boxed{\text{The west end of the wire is at the higher electrical potential.}}$$

Given:
- Initial current: $I_1 = 5.0\,\text{A}$
- Final current: $I_2 = 0.0\,\text{A}$
- Time interval: $\Delta t = 0.1\,\text{s}$
- Induced emf: $\varepsilon = 200\,\text{V}$

Concept/Formula used:

The self-induced emf is given by:
$$\varepsilon = -L\frac{dI}{dt}$$

For average values:
$$|\varepsilon| = L \left|\frac{\Delta I}{\Delta t}\right|$$

Calculation:

$$\left|\frac{\Delta I}{\Delta t}\right| = \frac{|I_2 - I_1|}{\Delta t} = \frac{|0 - 5.0|}{0.1} = \frac{5.0}{0.1} = 50\,\text{A s}^{-1}$$

$$L = \frac{|\varepsilon|}{|\Delta I / \Delta t|} = \frac{200}{50}$$

$$\boxed{L = 4\,\text{H}}$$

The self-inductance of the circuit is 4 H.

Given:
- Mutual inductance: $M = 1.5\,\text{H}$
- Change in current in coil 2: $\Delta I_2 = 20 - 0 = 20\,\text{A}$
- Time interval: $\Delta t = 0.5\,\text{s}$

Concept/Formula used:

The flux linkage with coil 1 due to current in coil 2 is:
$$N_1 \Phi_1 = M \cdot I_2$$

Therefore, the change in flux linkage with coil 1 is:
$$\Delta(N_1 \Phi_1) = M \cdot \Delta I_2$$

Calculation:

$$\Delta(N_1 \Phi_1) = 1.5 \times 20$$

$$\boxed{\Delta(N_1 \Phi_1) = 30\,\text{Wb}}$$

The change of flux linkage with the other coil is 30 Wb (weber-turns).

*Note:* The rate of change of flux linkage (i.e., the induced emf) would be:
$$\varepsilon = -M\frac{\Delta I_2}{\Delta t} = -1.5 \times \frac{20}{0.5} = -60\,\text{V}$$
but the question asks only for the change in flux linkage, which is $30\,\text{Wb}$.