HARD
JEE Main
IMPORTANT
Earn 100

Consider the situation shown in figure. The wires P1Q1 and P2Q2 are made to slide on the rails with the same speed 5 cm s-1. Find the electric current in the 19 Ω resistor if a both the wires move towards right and b If P1Q1 moves towards left but P2Q2 moves towards the right.

Question Image

Suppose the 19 Ω resistor of the previous problem (described above) is disconnected. Find the current through P2Q2 in the two situations a and b of that problem.

Important Points to Remember in Chapter -1 - Electromagnetic Induction from H C Verma CONCEPTS OF PHYSICS [VOLUME 2] Solutions

1. Magnetic fluxϕ:

(i) Magnetic flux linking with the coil of area A, when placed in a uniform magnetic field B such that the area vector makes an angle θ with the magnetic field is
ϕ=B.A=BAcosθ
Question Image
(ii) ϕ=BdA for non-uniform B

(iii) S.I. unit: T m2 or weber(Wb)

(iv) Dimensional formula: ML2T-2A-1
(v) It is a scalar quantity.

2. Faraday's law of electromagnetic induction:

(i) An emf is induced whenever the magnetic flux linking with the circuit changes.

(ii) The magnitude of the induced emf in any circuit is proportional to the rate of change of the magnetic flux linking the circuit, εdϕcoildt

(iii) If the coil has N turns, then the emf induced is ε=-N dϕdt, where ϕ is the flux linking with each turn of the coil.

(iv) ε=-NAcosθdBdt, when the only magnetic field B is variable.

(v) ε=-NBcosθdAdt, when the only area A is variable.

(vi) ε=-NBAdcosθdt=BAωsinωt, when only θ is variable.

(vii) Maximum emf induced is, εmax=NBAω

(viii) If the coil rotates about its own axis in a perpendicular magnetic field, then the emf induced is zero.

Question Image

3. Lenz's law:

(i) The direction of an induced emf is always such as to oppose the cause producing it.

(ii) Law of EMIεind=-dϕcoildt

(iii) The negative sign indicates that the induced emf opposes the change of the flux.

(iv) Induced current due to change in the field:

(a) If magnetic field strength increases with time as shown in the below figure, then the current induces in the anti-clockwise direction.

Question Image

(b) If magnetic field strength decreases with time, then the current induces in the clockwise direction.

(v) Induced current due to change in the area:

(a) If the area increases, then the flux linkage increases, and hence current induces in the anti-clockwise direction to oppose the change in flux.

Question Image

(b) If the area decreases, then the current induces in the clockwise direction.

4. Chargeq flowing in terms of change in fluxϕ:

q=1Rϕ, where R is the resistance of the coil.

5. Induced EMF and current:

(i) Changing the magnetic field will produce a voltage in a coil, causing a current to flow.

(ii) To be completely accurate, if the magnetic flux through a coil is changed, a voltage will be produced. This voltage is known as the induced emf & the current flowing (if the loop is complete) is called induced current.

6. Motional emf:
(i) Emf in vector form is ε=abv×B.dl=abB×dl.v=abdl×v.B

Question Image

(ii) ε=Blvsinθ, where B=flux density, l=length of the conductor, v=velocity of the conductor, and θ=angle between the direction of motion of conductor & field B.

Question Image

(iii) If the angle between v & B or v & l or l & B is 0°, then the emf induced is zero.

(iv) Maximum emf induced is εmax=Blv at θ=90°

7. Energy considerations when the rod is moving in a uniform magnetic field:

Consider a rod of length l is moving with a velocity v between the parallel rails placed in a uniform, perpendicular magnetic field B. A resistance R is connected to the ends of the rails as shown in the figure. Then, 
Question Image

(i) Emf induced in the rod: ε=Blv

(ii) Current flowing through the circuit: i=BlvR

(iii) Power dissipated in the resistor: Pdissipated=B2l2v2R

(iv) Mechanical power supplied to the rod: Pmech=Fmv=B2l2v2R

8. EMF induced in a rod rotating perpendicular to the magnetic field:

Question Image

(a) When a rod of length l is rotating in a uniform perpendicular magnetic field B about one of its ends with an angular velocity ω, the emf induced in the rod is ε=12Bωl2

(b) The direction of emf induced is decided by Fleming’s right-hand rule.

(c) Potential: VP>VQ

(ii) If a wheel of n spokes each of length l is rotating in a perpendicular magnetic field, then emf between centre of the wheel and its perimeter is, ε=12Bωl2

Question Image

(iii) If the rod is rotating about a point at distance l4 from one end, then

Question Image

VO-VP=+12Bl42ω, and VO-VQ=+12B3l42ω

VP-VQ=+14Bl2ω

9. Disc rotation in a magnetic field such that the axis of rotation is perpendicular to the magnetic field:

Consider a circular disc of radius R is rotating with an angular velocity ω in a uniform magnetic field of induction B. Then
Question Image

Emf induced between the centre and perimeter, ε=12BR2ω

10. Self and Mutual Induction:

(i) Self-induction:

When a current flowing through a coil is changed the flux linking with its own winding changes & due to the change in linking flux with the coil an emf is induced which is known as self-induced emf & this phenomenon is known as self-induction. This induced emf opposes the cause of Induction.

(ii) Self-inductance:

(a) The property of the coil or the circuit due to which it opposes any change of the current coil or the circuit is known as Self-Inductance. It is also called electrical inertia. Its S.I. unit is henry (H).

(b) Let ϕ be the flux linking with each turn, and N be the number of turns in the coil, then the total flux linkage is, Nϕ=Li

(c) Coefficient of self-inductance, L=ϕs for i=1 A, where i=current in the circuit, ϕs= magnetic flux linked with the circuit due to the current i.

(d) L depends only on (i) the shape of the loop & (ii) medium.

(e) self-induced emf, es=dϕnetdt=-ddtLi=-Ldidt (if L is constant)

(f) If the current is flowing from P to Q and increasing with time, then VP>VQ

Question Image

(g) If the current is flowing from P to Q and increasing with time, then VP<VQ

Question Image

(h) Self-inductance of a solenoid, L=μ0n2AI, where A=area of cross-section of the solenoid, μ=magnetic permeability of the core material, n=number of turns in the solenoid per unit length, and i=current in the solenoid.

Question Image

(i) Self-inductance of a circular coil carrying current i, having radius r and number of turns n is given by

L=μ0n2πr2

(j) Self-inductance of a toroidal solenoid:
Consider a toroidal coil of internal radius a, and external radius b with a thickness h having N turns. The self-inductance of this coil is

Question Image

 L=μ0N2h2πlnba

(iii) Mutual Induction:

(a) If two electric circuits are such that the magnetic field due to a current in one is partly or wholly linked with the other, the two coils are said to be electromagnetically coupled circuits.

(b) Any change of current in one produces a change of magnetic flux in the other & the latter opposes the change by inducing an emf within itself. This phenomenon is called mutual induction.

(c) The induced emf in the later circuit due to a change of current in the former is called mutually induced emf. The circuit in which the current is changed is called the primary & the other circuit in which the emf is induced is called the secondary.

(iv) Mutual-Inductance:

(a) The coefficient of mutual induction (mutual inductance) between two electromagnetically coupled circuits is the magnetic flux linked with the secondary per unit current in the primary.

(b) Mutual inductance=M=ϕmip=flux linked with secondary  current in the primary 

(c) mutually induced emf, Em=dϕmdt=-ddtMip=-Mdipdt(where M is constant)

(d) M depends on the geometry of loops, medium, orientation & distance between the loops.

(e) If two coils of self-inductances L1 and L2 are wound over each other, then the mutual inductance M=KL1L2, where K is called coupling constant. K lies between 0 and 1.

(f) Series combination L=L1+L2+, i same, V in the ratio of inductance, U in the ratio of inductance, ϕ in the ratio of inductance.

Question Image

(g) For two coils wound in the same direction connected in series and mutually interacting with each other,

Leq=L1+L2+2M

Question Image

(h) For two coils wound in opposite direction connected in series and mutually interacting with each other,

Leq=L1+L2-2M

Question Image

(i) For a parallel combination of inductors:

Question Image
Leq=L1L2L1+L2

(j) Two coils in parallel and mutually interacting with each other,

Question Image

Leq=L1L2-M2L1+L2±2M

(k) Mutual inductance between two co-axial solenoids,

Question Image

M=μ0n1n2πr12l

(l) Mutual inductance between two co-axial circular coils separated by some distance:

Question Image

M=μ0πr2R22x2+R23/2

(m) Mutual inductance between the infinite straight wire and a rectangular coil separated by some distance is,

Question Image

M=μ0b2πln1+al

11. Induced electric field:

(i) An emf is induced in a closed loop where magnetic flux is varied.

(ii) The induced electric field is not a conservative field because, for the induced electric field, the line integral E.dl around a closed path is non-zero.

12. Transformer:

Question Image

(i) For an ideal transformer,  ESEP=IPIS=NSNP

(ii) Efficiency,  η=Pout Pin ×100 %

13. Eddy Current:

(i) An eddy current is a current set up in a conductor in response to a changing magnetic field.

(ii) Eddy currents flow in closed loops in a plane perpendicular to the magnetic field.

(iii) By Lenz’s law, the current swirls in such a way as to create a magnetic field opposing the change.

(iv) For this to occur in a conductor, electrons swirl in a plane perpendicular to the magnetic field.

(v) Because of the tendency of eddy currents to oppose, eddy currents cause a loss of energy.

(vi) Eddy currents transform more useful forms of energy, such as kinetic energy, into heat, which is not generally useful.

Question Image

14. Super conduction loop in a magnetic field:

(i) Reistance R=0. So, emf induced ε=iR=i(0)=0. Therefore, ϕtotal = constant.

(ii) Thus, in a superconducting loop flux never changes, or it opposes 100%.

15. Magnetic energy stored in an inductor:

Energy stored in an inductor of inductance L and carrying a current i is, 
W=12Li2=12ϕi

16. Magnetic energy density:

Magnetic energy store per unit volume in terms of magnetic field induction is
u=12B2μ0μr

17. The energy between two interacting coils:

The total energy stored in two coils with inductances L1, L2 and mutual inductance M, if they carry currents I1 and I2 respectively is,

U=12L1I12+12L2I22+MI1I2

18. LR DC circuit:

(i) Growth of a current in an LR circuit:

Question Image

(a) Current at any time t is I=ER1-e-RtL If initial current =0

(b) The time constant of the circuitLR

(c) Final steady current I0=ER

(d) Inductor behaves as an open circuit at t=0 [If i=0]

(e) Inductor behaves as a short circuit at t= always.

(ii) Decay of Current:

(a) Initial current through the inductor =I0

(b) Current at any instant, i=I0e-RtL
Question Image
19. Alternating currents:

(i) A current that changes its direction periodically is called alternating current (AC). If a current maintains its direction constant it is called direct current (DC).

Question Image

(ii) Average value: Iavg=0TIdt0Tdt=1T0τIdt

(iii) RMS value: Irms=0TI2dt0Tdt

(iv) For a sinusoidal voltage V=V0sinωt: Vavg=2 V0π & Vrms=V02

(v) For a sinusoidal current I=I0sinωt+ϕ: lavg=2I0π & Irms=I02

20. AC generator:

An AC generator is an electric generator that converts mechanical energy into electrical energy in form of an alternative emf or alternating current. AC generators work on the principle of electromagnetic induction.

Question Image

21. AC circuit:

Question Image
Question Image
PBrush

(i) In a pure resistor, current and voltage are in phase.

(ii) In a pure capacitor, the current leads the voltage by 90°

(iii) In a pure inductor, the current lags the voltage by 90°

22. Reactance and Impedance:

(i) ωL is called inductive reactance and is denoted by XL

(ii) 1ωC is called capacitive reactance and is denoted by XC

23. Series LCR Circuit:

Consider a series LCR series circuit with resistance R, inductive reactance XL, and capacitive reactance XC.
(i) Impedance, Z=R2+XL-XC2

Question Image

(ii) Applied voltage in terms of voltages across the components, V=VR2+VL-VC2

24. Power in AC Circuits and Wattless current:

(i) Average power consumed in a cycle =02πωPdt2πω=12VmImcosφ

Question Image

(ii)  Pavg=Vm2Im2cosφ=Vrms Irms cosφ=Vrms2R, here cosφ is called a power factor.

(iii) The current in AC circuit is said to be wattless current when the average power consumed in such circuit corresponding to zero such current is also called idle current.

25. Wattless current:

In an AC circuit, if the average power consumed is zero, then the current is called a wattless current. Its formula is Irmssinϕ, where ϕ is the phase difference between current and voltage applied.

26. Resonance:

(i) At resonance: XL=XcZ=R,V=VR

(ii) Resonant frequency is fr=12πLC, where L is the inductance, and C is capacitance.

(iii) Nature of the circuit is resistive.

(iv) Current in the circuit is maximum, i.e., i0=V0R

(v) Power consumed at resonance is Pavg=Vrms2R

27. Bandwidth:

The bandwidth of RLC series circuit of resistance R, inductance L, and Capacitance C is
BW=RL

Question Image

As the resistance increases, peak current decreases, and the bandwidth increases.

28. Quality factor:

Q factor (also known as a Quality factor or Q-factor) is defined as a dimensionless parameter that describes the underdamped condition of an oscillator or resonator.
The quality factor of series RLC circuit of resistance R, inductance L, and Capacitance C is

Q=frBW=XL or XCR=1RLC

29. LC oscillations:

(i)   q=q0sinωt+θ, I=I0cosωt+θ, where I0=q0ω

Question Image

(ii) Energy =12Li2+q22C=q022C=12Li02=constant

(iii) Comparison of LC oscillations with SHM of spring:  qx, Iv, Lm, C1K

30. Comparison of damped mechanical & electrical systems:

(i) Series LCR circuit:
Consider a resistance R, inductnace L, and capacitance C are connected in series. Let q be the charge on the capacitor at time t. Then,

Question Image

d2qdt2+RLdqdt+1LCq=V0Lcosωt

compare with the mechanical damped system equation,

d2xdt2+bmdxdt+kmx=F0mcosωt, where b=damping coefficient.

Mechanical system Electrical systems (series RLC)
Displacement x Charge q
Driving force F Driving voltage v
Kinetic energy 12mv2 Electromagnetic energy of moving charge 12Ldqdt2=12Li2
Potential energy 12kx2 The energy of the static charge q22C
Mass m Inductance L
Power P=Fv Power P=Vi
Damping b Resistance R
Spring constant k 1C

(ii) Parallel LCR circuit:

(a) Consider a resistance R, inductnace L, and capacitance C are connected in parallel. Let q be the charge on the capacitor at time t. Then,

Question Image

I=IL+IC+IR=ϕL+ddtCdϕdt+1Rdϕdtd2ϕdt2+1RCdϕdt+1LCϕ=V0Zsinωt

(b) Displacement x Flux linkage ϕ

(c) Velocity dxdtVoltage dϕdt

(d) Mass m Capacitance C

(e) Spring constant k Reciprocal Inductance 1L

(f) Damping coefficient b Reciprocal resistance 1R; Driving force F Current i