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Correct option is: (A) 0.1 V

Data : L = 4.7 × 10^-2 H, 

I_i = 15A, I_i = -15 A, 

∆f = 0.01 s

The rate of change of current,

 \(\cfrac{dI}{dt}\) = \(\cfrac {I_f−I_i}{Δt}\) = \(\cfrac{(−15)-15}{0.01}\) = – 3000 A/s

∴ The maximum self-induced emf,

e = – L \(\cfrac{dI}{dt}\) (4.7 × 10^-2) (- 3000) = 141 V

(A) anticlockwise

Anticlockwise/adcba.

(a) The values displayed by, Ashok are :

(i) presence of mind, 

(ii) higher degree of general awareness, and 

(iii) convincing capacity. 

(b) When the current was passed through, the magnetic flux linked through the bird changed and so induced current flew in the birds body. Its wings due to opposite current experienced mutual repulsion and the bird flew away.

The electric current flowing in a wire in the direction Clockwise.

(i) Concern for the nation, cost economic attitude, and sharing knowledge .

(ii) ε = -L dI /dt = 233.3 V

A has positive polarity.

(i) Clockwise in 1

(ii) Anticlockwise in 2

(i) Respecting law and obeying the rules enacted by the nation; self discipline and avoiding arguments.

(ii) The change in the magnetic field in a metal detector due to the metals carried by the contents of checking which, induces the change in current in the coil of the metal detector. This detection of change sets up an alarm.

Number of turns in rectangular coil : (N) = 1000
Magnetic field (B) = 0.2 T
Rotation frequency (f) = 4200 rpm
∴ Angular velocity = ω = \(\frac{2 \pi f}{t}\)
= \(\frac{2 \times \pi \times 4200}{60}\) = 439.60 rad/sec
∴ Maximum induced emf.
ε₀ = NABω
= 1000 × 0.2 × 0.02 × 439.60
= 1758.4 volt

Due to the change in magnetic flux, the eddy currents are produced in a metallic wire. So an opposing force is experienced while pulling or pushing a metallic plate in a uniform magnetic field.

(c) 20 Bx² cosωt
Given N = 20, A = x then
ϕ = NBA cos ωt = 20Bx² cos ωt

Zero, due to the no change in magnetic flux.

(a) Charge
\(\frac{\phi}{R}\) = Charge (q)

(b) Zero
ϕ = BA cos 90° = 0
∴ Induced emf = 0

The induced e.m.f. is given by coil:
ε = ε₀ sinωt = NABω sinωt
Where N = Number of turn, A = Area of cross section, B = Magnetic field, ω = Angular velocity.

(d) Resistance of wire in coil
Resistance of wire

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

The SI unit of self inductance or coefficient of self induction or inductance as it is commonly called is called the henry (H).

The self-inductance of a coil is 1 henry, if an emf of 1 volt is induced in the coil when the current through the same coil changes at the rate of 1 ampere per second.

The dimensions of self inductance or coefficient of self induction are [ML²T^-2 I^-2 ].

1 henry = 1 H = 1 V/A.s = 1 T.m² /A

[ Note : The unit henry is named in honour of Joseph Henry (1797-1878) US physicist.]

(i) Magnetic induction, B : 

SI unit : the tesla (T) : 1 T = 1 Wb / m² 

Dimensions: [B] = [MT^-2 I^-1 ].

(ii) Magnetic flux, Φ_m : 

SI unit : the weber (Wb)

Dimensions : [Φ_m] = [B][A] 

= [MT^-2 I^-1 ][L² ] = [ML ²T^-2I^-1 ]

If the transformer converts an alternating current with low voltage into an alternating current with high voltage, it is called step-up transformer. On the contrary, if the transformer converts alternating current with high voltage into an alternating current with low voltage, then it is called step-down transformer.

1. The current is drawn directly from fixed terminals on the stator without the use of brush contacts. 

2. The insulation of stationary armature winding is easier. 

3. The number of sliding contacts (slip rings) is reduced. Moreover, the sliding contacts are used for low-voltage DC Source.

4. Armature windings can be constructed more rigidly to prevent deformation due to any mechanical stress.

Consider a narrow air-cored toroid of circular cross section of radius r, central radius R and number of turns N. So that, assuming r << R, the magnetic field in the toroidal cavity is considered to be uniform, equal to

B = \(\cfrac{μ_0 NI}{2πR}\) = µ₀nI ………….. (1)

where n = \(\cfrac{N}{2πR}\) is the number of turns of the wire 2nR per unit length. The area of cross section, A = πr².

The magnetic flux through one turn is

Φ_m = BA = µ₀nIA ………… (2)

Hence, the self inductance of the toroid,

L = \(\cfrac{NΦ_m}I\) = (2πRn) µ nA = µ₀2πRn² A = µ₀n² V …………… (3)

= \(\cfrac{μ_0N^2r^2}{2R}\) ………….. (4)

where V = 2πRA is the volume of the toroidal cavity. Equation (3) or (4) gives the required expression.

Three-phase system has many advantages over single-phase system, which is as follows: 

(i) For a given dimension of the generator, three-phase machine produces higher power output than a single-phase machine. 

(ii) For the same capacity, three-phase alternator is smaller in size when compared to single phase alternator. 

(iii) Three-phase transmission system is cheaper. A relatively thinner wire is sufficient for transmission of three-phase power

The maximum value of the alternating emf induced in the secondary coil of a transformer depends on

1. the ratio of the number of turns of the secondary coil to that of the primary coil 

2. the maximum value of the alternating emf applied to the primary coil 

3. the core of the transformer.

\(\cfrac{V_S}{V_P}\) = \(\cfrac{N_S}{N_P}\)

\(\therefore\) \(\cfrac{V_S}{100\,V}\) = \(\cfrac{200}{100}\)

\(\therefore\) V_s = 200 V is the required emf.

Data : N_p = 20, N_s = 300, V_p = 10 sin ωt V

V_s = \(\cfrac{N_s}{N_p}\) V_p

= \(\cfrac{300}{20}\) x 10 sin ωt

= 150 sin ωt V

This is of the form V₀ sin ωt, where V₀ is the peak (or maximum) voltage. 

∴ The maximum voltage in the secondary coil is 150 V.

Data : R = 3 Ω, N_s = 60, N_p = 1200, V_p = 240 V

Vs = \(\cfrac{N_S}{N_P}\) × V_p

= \(\cfrac{60}{1200}\) x 240 = 12 V (peak)

∴ The peak value of the current in the resistor in the transformer secondary coil is

I_s = \(\cfrac{V_S}{R}\) = \(\cfrac{12}{3}\) = 4 A

Clockwise on the side of the observer.

Q-factor measures the selectivity in resonant circuit

Angular resonant frequency (co) is \(\frac{1}{\sqrt{LC}}\).

When the frequency of the applied alternating source (ω_r ) is equal to the natural frequency \([\frac{1}{\sqrt{LC}}]\) of the RLC circuit, the current in the circuit reaches its maximum value. Then the circuit is said to be in electrical resonance. The frequency at which resonance takes place is called resonant frequency. Resonant angular frequency, ω_{r = \(\frac{1}{\sqrt{LC}}\)}

At resonance, the impedance of the circuit is equal to the resistance in the circuit,
i.e. Z = R
∴ power factor of the circuit,
cos φ = \(\frac{R}{Z}= \frac{R}{R}= 1\)

Data : L = 3 H, R = 100 Ω, I = 2 A 

(a) Magnetic energy stored,

U_m = \(\cfrac12\)LI² = \(\cfrac12\)(3) (2)² = 6 J

(b) Power dissipated in the resistance of the coil,

P = I²R = (2)² (100) = 400 W

Lenz’s Law is in accordance with the law of conservation of energy.

Conservation of energy: 

The truth of Lenz’s law can be established on the basis of the law of conservation of energy. According to Lenz’s law, when a magnet is moved either towards or away from a coil, the induced current produced opposes its motion. As a result, there will always be a resisting force on the moving magnet. 

Work has to be done by some external agency to move the magnet against this resisting force. Here the mechanical energy of the moving magnet is converted into the electrical energy which in turn, gets converted into Joule heat in the coil i.e., energy is converted from one form to another.

Yes, Lenz’s law consistent with the law of conservation of energy.

The answer is (D) Energy.

(i) Induced emf, ε = - \(\frac{dφ}{dt}\) = -\(\frac{d}{dt}\) (BA cos ωt)

= BA ω sin ωt

As B, A, ω are same for both loops, so induced emf is same in both loops. 

(ii) Current induced, I = \(\frac{ε}{R}\) = \(\frac{ε}{pl/A}\) = \(\frac{εA}{pl}\)

As area A, length l and emf ε are same for both loops but resistivity ρ is less for copper, therefore current I induced is larger in copper loop.

(a) In Fig. (i) the rectangular loop abcd and in Fig. (iii) circular loop are entering the magnetic field, so the flux linked with them increases; The direction of induced currents in these coils, will be such as to oppose the increase of magnetic flux; hence the magnetic field due to current induced will be upward, i.e., currents induced will flow anticlockwise. 

(b) In Fig. (ii), the triangular loop abc and infig. (iv) The zig-zag shaped loop are emerging from the magnetic field, therefore magnetic flux linked with these loops decreases. The currents induced in them will tend to increase the magnetic field in downward direction, so the currents will flow clockwise. 

Thus in fig. (i) Current flows anticlockwise, 

In fig. (ii) Current flows clockwise, 

In fig. (iii) Current flows anticlockwise, 

In fig. (iv) Current flows clockwise.

(a) When the tapping key is just closed, the current produced in the left loop flows clockwise, so magnetic field induced will flow along negative axis; the current induced in right coil will oppose the magnetic field produced, so current in right coil will flow anticlockwise, i.e., direction of current will be along yzx. 

(b) The current in coil is anticlockwise. When rheostat setting is being changed, the resistance of the right circuit is decreasing, so current is increasing, the current induced in left loop will oppose the increase of current, so current induced in left coil will flow clockwise i.e., along zyx. 

(c) Induced current in the right coil is along xry. 

(d) No induced current because magnetic field lines lie in the plane of loop.

(a) The direction of current is along qrpq because the current induced in solenoid will oppose the approach of magnet, so from looking on magnet side, the current at nearer face should flow clockwise. 

(b) In this case the current induced in coil pq will oppose the approach of magnet while coil xy will oppose the recession of magnet; so nearer faces of coils will act as S-poles. Accordingly the direction of current in coil pq will be along qrp and in coil xy it will be along yzx.

(i) For the given periphery the area of a circle is maximum. When a coil takes a circular shape, the magnetic flux linked with coil increases, so current induced in the coil will tend to decrease the flux and so will produce a magnetic field upward. As a result the current induced in the coil will flow anticlockwise i.e., along adcb. 

(ii) For given periphery the area of circle is maximum. When circular coil takes the shape of narrow straight wire, the magnetic flux linked with the coil decreases, so current induced in the coil will tend to oppose the decrease in magnetic flux; hence it will produce upward magnetic field, so current induced in the coil will flow anticlockwise i.e., along a′ b′ c′ b′.

In Figure (a): When a wire of irregular shape turns into a circular (loop the magnetic flux linked with the loop increases due to increase in area) The irregular shape. The induced e.m.f. will cause current to flow in such a direction, so that the wire forming the loop is pulled in ward from all sides. It requires that current should flow.

In figure (b): By applying Lenz’s Law, it follows that the current will flow in the direction a, d, c, b, a.

(A) maintain the original magnetic flux through the loop

Due to the motion of coil, the magnetic flux linked with the coil increases. So by Lenz’s law, the current induced in the coil will oppose this increase, hence tend to produce a field upward, so current induced in the coil will flow anticlockwise.

i.e., along PSRQP

So far the loop remains in the magnetic field, there is no change in magnetic flux linked with the loop and so no current will be induced in it, but when the loop comes out of the magnetic field, the flux linked with it will decrease and so the current will be induced so as to oppose the decrease in magnetic flux, i.e., it will cause magnetic field downwards; so the direction of current will be clockwise.

As there is no change in magnetic flux, so no current is induced in the loop.