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An AC generator is which converts mechanical energy into electrical energy (alternating emf). It works on the principle of electromagnetic induction. 

Expression for Induced emf: We know that if a charge q moves with velocity \(\vec{V}\) in a magnetic field of strength \(\vec{B}\), making an angle θ then magnetic Lorentz force

F = q vB sin θ

If \(\vec{V}\) and  \(\vec{B}\) mutually perpendicular, then θ = 90°

F = q vB sin 90° = qvB

The direction of this force is perpendicular to both \(\vec{V}\) and \(\vec{B}\) and is given by Fleming’s left hand rule.

Suppose a thin conducting rod PQ is placed on two parallel metallic rails CD and MN in a magnetic field of strength \(\vec{B}\). The direction of magnetic field \(\vec{B}\) is perpendicular to the plane of paper, downward. In fig \(\vec{B}\) is represented by cross (×) marks. Suppose the rod is moving with velocity \(\vec{V}\), perpendicular to its own length, towards the right. We know that metallic conductors contain free electrons, which can move within the metal. As charge on electron, q = – e therefore, each electron experiences a magnetic Lorentz force, F_m = evB, whose direction, according to Fleming’s left hand rule, will be from P to Q Thus the electrons are displaced from end P towards end Q Consequently the end P of rod becomes positively charged and end Q negatively charged. Thus a potential difference is produced between the ends of the conductor. This is the induced emf.

Due to induced emf, an electric field is produced in the conducting rod. The strength of this electric field

E = \(\frac{V}{l}\) ....(i)

And its direction is from (+) to (–) charge, i.e., from P to Q. 

The force on a free electron due to this electric field, F_e = _eE ...(ii) 

The direction of this force is from Q to P which is opposite to that of electric field. Thus the emf produced opposes the motion of electrons caused due to Lorentz force. This is in accordance with Lenz’s law. As the number of electrons at end becomes more and more, the magnitude of electric force Fe goes on increasing, and a stage comes when electric force \(\vec{Fe}\) and magnetic force \(\vec{Fm}\) become equal and opposite. In this situation the potential difference produced across the ends of rod becomes constant. In this condition 

F_e = F_m

_eE = _evB or E = B_v ...(iii)

∴ The potential difference produced,

V = EI = B vI Volt

Also the induced current I = \(\frac{V}{R}\) = \(\frac{Bvl}{R}\) ampere

Consider an inductor of sell inductance L connected in a circuit When the circuit is dosed, the current in the circuit increases and so does the magnetic flux linked with the coiL At any instant the magnitude of the induced emf is

e = L \(\cfrac{di}{dt}\)

The power consumed in the inductor is 

P = ei = L \(\cfrac{di}{dt}\) . i

[Alternatively, the work done in moving a charge dq against this emf e is 

dw = edq = L \(\cfrac{di}{dt}\). dq = Li ∙ di (∵ \(\cfrac{dq}{dt}\) = i)

This work done is stored in the magnetic field of the inductor. dw = du.]

The total energy stored In the magnetic field when the current increases from 0 to I In a time interval from 0 to t can be determined by integrating this expression :

U_m = ∫₀^t Pdt = ∫₀^I Lidi = L ∫₀^I idi = \(\cfrac12\) LI²

which is the required expression for the stored magnetic energy.

[Note: Compare this with the electric energy stored in a capacitor, U_e = \(\cfrac12\) CV²]

By inserting a ferromagnetic material inside the coils. 

Eddy currents are used to advantage in certain applications like: 

(i) Magnetic braking in trains: Strong electromagnets are situated above the rails in some electrically powered trains. When the electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train. As there are no mechanical linkages, the braking effect is smooth. 

(ii) Electromagnetic damping: Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly. 

(iii) Induction furnace: Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil which surrounds the metals to be melted. The eddy currents generated in the metals produce high temperatures sufficient to melt it. 

(iv) Electric power meters: The shiny metal disc in the electric power meter (analogue type) rotates due to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil.  

When the current through a coil goes on changing, the magnetic flux linked with the coil also goes on changing. The magnetic flux (NΦ ) linked with the coil at any instant is directly proportional to the current (I) through the coil at that instant.

NΦ_m ∝ I

∴ NΦ_m = LI

where L is a constant, dependent on the geometry of the coil, called the self inductance or the coefficient, of self induction of the coil. The self-induced emf in the coil is

e = - \(\cfrac{dNϕ _m}{dt}\) = - \(\cfrac{d}{dt}\) (LI) = - L \(\cfrac{DI}{dt}\)

In magnitude,

e = L \(\cfrac{d}{dt}\)

\(\therefore\) L = \(\cfrac{e}{dI/dt}\)

Definition : The self inductance or the coefficient of self induction of a coil is defined as the emf induced in the coil per unit time rate of change of current in the same coil. OR (using L = NΦ_m /I), the self inductance of a coil is the ratio of magnetic flux linked with the coil to the current in it.

Magnetic flux linked with a coil

ϕ=NBA cosθ

Since the magnetic field B is parallel to the plane of area A, i.e., θ=90°

So,  ϕ=0

Correct option is: (A) BLv

Correct option is: (B) 100 Wb

When the current in the solenoid decreases a current flows in the same direction in the metal ring as in the solenoid. Thus there will be a downward force. This means the ring will remain on the cardboard. The upward reaction of the cardboard on the ring will increase.

(i) When the magnet is rotated about its own axis, then due to symmetry of magnet the magnetic flux linked with circular coil remains unchanged, hence no emf is induced at terminals of coil. 

(ii) When the magnet is rotated about an axis perpendicular to the length, the positions of N and S poles of magnet changes continuously; so the magnetic flux linked with the coil changes continuously; hence the emf is induced at the terminals of the coil.

Induced emf can be produced by changing magnetic flux in any of the following ways: 

1. By changing the magnetic field B 

2. By changing the area A of the coil and

3. By changing the relative orientation 0 of the coil with magnetic field.

Inductor is a device used to store energy in a magnetic field when an electric current flows through it. The typical examples are coils, solenoids and toroids

Self – inductance is the ratio of the magnetic flux linked with a coil to the current flowing through it. 

If the magnetic flux is changed by changing the current, an emf is induced in that same coil. This phenomenon is known as self-induction

Self-inductance of a coil is its property virture of which the coil opposes any change in the current flowing through it.

This property of a coil is analogous to mechanical inertia. That is why self-induction is called the inertia of electricity.

(b) l decreases and A increases. 

Data: N = 1000, D = 0.1 cm, L = 2.4π × 10^-5H

I = 3A, μ₀ = 4π × 10^-7 H/m

The number of turns per unit length.

n = \(\cfrac1{1\,mm}\) = 1 mm^-1 = 10³ m^-1

and the length of the solenoid,

l = ND = 1000 × 0.1 = 100 cm = 1 m

L = μ₀n² lA

(a) The area of cross section 

A = \(\cfrac{L}{μ_ 0n^2l}\) = \(\cfrac{2.4π×10^{-5}}{(4π×10^{-7} )(10^3)^2 (1)}\) = \(\cfrac{24π}{4π}\) x 10^-5

= 6 × 10^-5 m²

(b) Magnetic flux through one turn, 

Φ_m = BA = (μ₀nI)A 

= (4π × 10^-7)(10³)(3)(6 × 10^-5) 

= 72π × 10^-9 Wb

Electromagnetic Induction 

(i) For larger deflection, coil P should be moved at a faster rate. 

(ii) Faraday law: The induced emf is directly proportional to rate of change of magnetic flux linked with the circuit.

Yes, provided the coil is in a closed circuit.

(1) The deflection in galvanometer may be made large by 

(i) Moving coil C₂ towards C₁ with high speed. 

(ii) By placing a soft iron laminated core at the centre of coil C₁. 

(2) The induced current can be demonstrated by connecting a torch bulb (in place of galvanometer) in Coil C₁. Due to induced current the bulb begins to glow

Law of conservation of energy is the basis of Lenz’s law of electromagnetic inductIon.

According to Lena’s law, the direction of the induced emf or current is such as to oppose the change that produces it. The change that induces a current may be

(i) the motion of a conductor in a magnetic field 

OR

(ii) the change of the magnetic flux through a stationary circuit. 

In the first case, the direction of induced emf in the moving conductor Is such that the direction of the side-thrust exerted on the conductor by the magnetic field is opposite in direction to its motion. The motion of the conductor is, therefore, opposed.

In the second case, the induced current sets up a magnetic field of its own which within the area bounded by the circuit is (a) opposite to the original magnetic field if this field is increasing, but (b) is in the same direction as the original field, if the field is decreasing. Thus, it is the change in magnetic flux through the circuit (not the flux itself) which is opposed by the induced current.

Suppose dΦ_m Is the change in the magnetic flux through a coil or circuit in time dt. Then, by Faraday’s second law of electromagnetic induction, the magnitude of the einf Induced is 

e ∝ \(\cfrac{dΦ_m}{dt}\) or e = k \(\cfrac{dΦ_m}{dt}\)

where dΦ_m/dt is the rate of change of magnetic flux

linked with the coil and k is a constant of proportionality. The Sl units of e (the volt) and dΦ df (the weber per second) are so selected that the constant of proportionality, k, becomes unity. Combining Faraday’s law and Lents law of electromagnetic induction, the induced emf

e = - \(\cfrac{dΦ_m}{dt}\)

where the minus sign is Included to indicate the polarity of the induced emf as given by Lents law. This polarity simply determines the direction of the induced current in a dosed loop. If a coil has N tightly wound loops, the induced emf will be N times greater than for a single loop, so that

e = – N \(\cfrac{dΦ_m}{dt}\)

where \(\cfrac{dΦ_m}{dt}\) is the rate of change of magnetic flux through one loop.

When the North-pole of a magnet is moved towards a coil connected to a galvanometer, the galvanometer in the circuit shows a deflection indicating a current (and hence an emf) in the circuit. The deflection continues as long as the magnet is in motion. A deflection can be observed if and only if the coil and the magnet are in relative motion. When the magnet is moved away from the coil, the galvanometer shows a deflection in the opposite direction. Bringing the South-pole towards the coil produces the opposite deflection as bringing the North pole. Faster the magnet or the coil is moved, larger is the deflection produced. By this experiment we can conclude that: the relative motion between the coil and the magnet generates an emf (current) in the coil. 

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(c) 4 B_o L² Wb. 

(a) If both assertion and reason are true and the reason in the correct explanation of the assertion.

Inductance coils made of copper will have very small ohmic resistance

Correct answe is (c) \(\frac{1}{2}\) Li²

(d) square of the number of turns

correct answer is (a) 5 mH

(c) N²

According to self inductance of long solenoid

L= \(\frac{μ_0N^2A}{l}\)

⇒ L ∝ N²

(b) Law of conservation of energy

(c) If assertion is true but reason is false

According to Faraday’s law, the conversion of mechanical energy into electrical energy is in accordance with the law of conservation of energy. It is also clearly known that in pure resistance, the emf is in phase with the current.

First law:

Whenever magnetic flux linked with a closed circuit changes, an emf is induced in the circuit. 

Second law: 

The magnitude of induced emf in a closed circuit is equal to the time rate of change of magnetic flux linked with the circuit

(i) → (c) 

(ii) → (a) 

(iii) → (d) 

(iv) → (b)

Data: R = 15Ω, 

v = 0.6 m/s, 

l = 0.25m, 

B = 0.35T 

(a) Induced emf, e = Blv = (0.35)(0.25)(0.6) 

= 0.0525 V = 52.5 mV

The current in the rod, I = \(\cfrac{e}R\) = \(\cfrac{52.5}{15}\) = 3 5 mA

(b) Power dissipated, P = eI = 0.0525 × 3.5 × 10^-4

= 0.184 mW

R = 25Ω

(a) e = 50V, T = 0.25s

i = e/R = 2A, H = i²RT

= 4 x 25 x 0.25 = 25J

(b) e = 50V, T = 0.25s

i = e/R = 2A, H = i²RT = 25J

(c) Since energy is a scalar quantity

Net thermal energy developed = 25J + 25J = 50J.

Data : M = 2 H, dI = – 4 A, dt = 2.5 × 10^-4 s

The induced emf, e = – M \(\cfrac{dI}{dt}\) = - \(\cfrac{2\times(-4)}{2.5\times10^{-4}}\)

= \(\cfrac{8}{2.5}\) x 10⁴ = 3.2 x 10⁴ V

For two inductively coupled coils, the fraction of the magnetic flux produced by the current in one coil (primary) that is linked with the other coil (secondary) is called the coefficient of magnetic coupling between the two coils.

The coupling coefficient K shows how good the coupling between the two coils is; 0 ≤ K ≤ 1. In the ideal case when all the flux of the primary passes through the secondary, K=l. For coils which are not coupled, K = 0. Two coils are tightly coupled if K > 0.5 and loosely coupled if K < 0.5.

[ Note ; For iron-core coupled circuits, the value of K may be as high as 0.99, for air-core coupled circuits, K varies between 0.4 to 0.8. ]

(D) the magnetic flux

An alternating emf VP from an ac source is applied across the primary coil of a transformer, shown in figure.

This sets up an alternating current fP in the primary circuit and also produces an alternating magnetic flux through the primary coil such that

V_p = -N_p \(\cfrac{dΦ_P}{dt}\)………….. (1)

where N_p is the number of turns of the primary coil and Φ_p is the magnetic flux through each turn.

Assuming an ideal transformer (i.e., there is no leakage of magnetic flux), the same magnetic flux links both the primary and the secondary

coils, i.e., Φ_p = Φ_s .

As a result, the alternating emf induced in the secondary coil,

V_s = – N_s \(\cfrac{dΦ_S}{dt}\) = - N_s \(\cfrac{dΦ_P}{dt}\)……………… (2)

where N_s is the number of turns of the secondary coil.

From Eqs. (1) and (2),

\(\cfrac{V_s}{V_p}\) = \(\cfrac{N_s}{N_p}\) or V_s = V_p \(\cfrac{N_s}{N_p}\)…………… (3)

Case (1) : If N_s > N_p , V_s > V_p . Then, the transformer is called a step-up transformer. 

Case (2) : If N_s < N_p , V_s < V_p . Then the transformer is called a step-down transformer

Ignoring power losses, the power delivered to the primary coil equals that taken out of the secondary coil, so that V_pI_p = V_sI_s …………. (4) 

From Eqs. (3) and (4),

\(\cfrac{I_p}{I_s}\) = \(\cfrac{V_s}{V_p}\) = \(\cfrac{N_s}{N_p}\)

Correct option is (B) 80

Data: V_p = 220V, V_s = 20V, P_s = 100W, I_p = 0.5 A

The Input power. P_p = I_pV_p = (0.5)220) = 110 W

The output power, P_s = 100 W

∴ The efficiency of the transformer

= \(\cfrac{output \,power}{input \,power}\) = \(\cfrac{100}{110}\) = 0.9091 or 90.91%

The ratio of the number of turns in the secondary coil (N_s) to that in the primary coil (N_p) is called the turns ratio of a transformer. The turns ratio \(\cfrac{N_s}{N_p}\) > 1 for a step-up transformer.

The turns ratio \(\cfrac{N_s}{N_p}\)< 1 for a step-down transformer.

Mutual inductance = 10^-2/2=5mH

Flux = 5 ×10^–3 ×1=5 ×10^–3Wb.

Data : N₁ = 100, N₂ = 200, L₁ = 25 mH, L₂ = 40 mH,

I₁ = 6 mA, dI /dt = 4 A/s

(a) The flux per unit turn in coil 1,

Φ₂₁ = \(\cfrac{L_1I_1}{N_1}\) = \(\cfrac{(25\times10^{-3})(6\times10^{-3})}{100}\)

= 1.5 × 10^-6 Wb =1.5 μ Wb

(b) The magnitude of the self induced emf in coil 1 is

L₁ = \(\cfrac{dI_1}{dt}\) = (25 × 10^-3)(4) = 0.1 V

(c) The flux per unit turn in coil 2,

Φ₂₁ = \(\cfrac{MI_1}{N_2}\) = \(\cfrac{(3\times10^{-3})(6\times10^{-3})}{200}\)

= 90 × 10^-9 Wb = 90 nWb

(d) The mutually induced emf in coil 2 is

e₂₁ = M \(\cfrac{dI_1}{dt}\) = (3 × 10^-3)(4) = 12 × 10^-3 V

= 12 mV

Induced emf, |ε| = \(\frac{dΦ}{dt} = A\frac{dB}{dt}\) = 8 x 2 x 10^-4 x 0.02

ε = 3.2 × 10^-5 V

Induced current, I = \(\frac{ε}{R}\) = 2 × 10^-5 A

Power loss = I²R = 4 × 10^-10 × 1.6 = 6.4 × 10^-10 W

When the coil is in vertical position, the angle between the normal to the plane of the coil and magnetic field is zero.