CBSE Class 12 Physics Electromagnetic Induction And Alternating Currents chapter notes and important questions

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

GIST

1 The phenomenon in which electric current is generated by varying magnetic fields is called electromagnetic induction.

2 Magnetic flux through a surface of area A placed in a uniform magnetic field B is defined as Φ_B= B.A = BACosθ where θ is the angle between B and A. B

3 Magnetic flux is a scalar quantity and its SI unit is weber (Wb). Its dimensional formula is [*Φ] = [ML²T^-2A-¹].

4 Faraday’s laws of induction states that the magnitude of the induced e.m.f in a circuit is equal to the time rate of change of magnitude flux through the circuit.ε=-dΦ_B/dt

5 According to Lenz law, the direction of induced current or the polarity of the induced e.m.f is such that it tends to oppose the change in magnetic flux that produces it. (The negative sign in Faraday’s law indicates this fact.)

6 Lenz law obeys the principle of energy conservation.

7 The induced e.m.f can be produced by changing the (i) magnitude of B (ii) area A (iii) angle θ between the direction of B and normal to the surface area A.

8 When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced e.m.f is called motional e.m.f produced across the ends of the rod which is given by ε = Blv.

9 Changing magnetic fields can setup current loops in nearby metal bodies (any conductor). Such currents are called eddy currents. They dissipate energy as heat which can be minimized by laminating the conductor.

10 Inductance is the ratio of the flux linkage to current.

11 When a current in a coil changes it induces a back e.m.f in the same coil. The selfinduced e.m.f is given by where L is the self-inductance of the coil. It is a measure of inertia of the coil against the change of current through it. Its S.I unit is henry (H).

12 A changing current in a coil can induce an e.m.f in a nearby coil. This relation, shows that Mutual inductance of coil 1 with respect to coil 2 (M₁₂ ) is due to change of current in coil 2. (M₁₂ = M₂₁ ). 

13 The self-inductance of a long solenoid is given by L = μ_o n² Al where A is the area of 0 crosssection of the solenoid, l is its length and n is the number of turns per unit length.

14 The mutual inductance of two co-axial coils is given by M₁₂ = M₂₁ = μ₀ n₁ n₂ Al where n₁ & n₂ are the number of turns per unit length of coils 1 & 2. A is the area of cross-  section and l is the length of the solenoids

15 Energy stored in an inductor in the form of magnetic field is  U_B=1/2Li² _maxand U_B=B²/ 2μ₀ Magnetic energy density.

16 In an A.C. generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction. Rotation of rectangular coil in a magnetic field causes change in flux (Φ = NBACosωt).

* Change in flux induces e.m.f in the coil which is given by ε= -dΦ/dt = NBAωSinωt ε = ε₀ Sinωt 

* Current induced in the coil I = ε/R = ε₀ Sinωt/R = I₀ Sinωt 

17 An alternating voltage ε=ε₀ Sinωt, applied to a resistor R drives a current I = I₀ Sinωt in the resistor, I₀ = ε₀ /R where ε₀ & I₀ are the peak values of voltage and current. (also represented by V_m & I_m )

Question. Figure shows a straight wire of length l current i. The magnitude of magnetic field produced by the current at point P is:

Answer : C

Question. A cell is connected between the points A and C of a circular conductor ABCD of centre 'O' with angle AOC = 60°, If B₁ and B₂ are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, the ratio B₁ / B₂ is:

a. 0.2
b. 6
c. 1
d. 5
Answer : C

Question. For the solenoid shown in figure. The magnetic field at point P is:

Answer : A

Question. A proton of energy 200 MeV enters the magnetic field of 5 T. If direction of field is from south to north and motion is upward, the force acting on it will be:
a. Zero
b. 1.6 10⁻¹⁰ N
c. 3.2 10⁻⁸ N 
d. 1.6 10⁻⁶ N
Answer : B

Question. A particle with 10–11 coulomb of charge and 10^–7 kg mass is moving with a velocity of 10⁸ m/s along the y-axis. A uniform static magnetic field B = 0.5 Tesla is acting along the x-direction. The force on the particle is:
a. 5 × 10^–11 N along iˆ
b. 5 × 10³ N along k ˆ
c. 5 × 10^–11 N along − ˆj
d. 5 × 10^–4 N along − k ˆ
Answer : D

Question. An electron is moving along positive x-axis. To get it moving on an anticlockwise circular path in x-y plane, a magnetic field is applied:
a. Along positive y-axis
b. Along positive z-axis
c. Along negative y-axis
d. Along negative z-axis
Answer : A

Question. The earth’s magnetic field at a given point is 0.5 10^-5 Wb-m^-2 . This field is to be annulled by magnetic induction at the center of a circular conducting loop of radius 5.0cm. The current required to be flown in the loop is nearly:
a. 0.2 A
b. 0.4A
c. 4A
d. 40A
Answer : B

Question. In the figure shown there are two semicircles of radii r₁ and r₂ in which a current i is flowing. The magnetic induction at the centre O will be:

Answer : C

Question. Two very long straight, particle wires carry steady currents i and – i respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires. It's instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is:
a. μ₀iqv / 2πd
b. μ₀iqv / πd
c. 2μ₀iqv / πd
d. Zero
Answer : D

Question. A metallic block carrying current i is subjected to a uniform magnetic induction B as shown in the figure. The moving charges experience a force F given by ……. which results in the lowering of the potential of the face ……. Assume the speed of the carriers to be v :

a. eVBkˆ, ABCD
b. eVBkˆ, ABCD
c. −eVBkˆ, ABCD
d. −eVBkˆ,EFGH
Answer : C

Question. A current carrying circular loop is freely suspended by a long thread. The plane of the loop will point in the direction:
a. Wherever left free
b. North-south
c. East-west
d. At 45° with the east-west direction
Answer : C

Question. A particle of charge 16 x 10⁻¹⁸ coulomb moving with velocity 10 m/s along the x-axis enters a region where a magnetic field of induction B is along the y-axis, and an electric field of magnitude 10⁴ V/m is along the negative zaxis. If the charged particle continuous moving along the x-axis, the magnitude of B is:
a. 10⁻³ Wb /m²
b. 10⁻³ Wb /m²
c. 10⁻⁵ Wb /m²
d. 10⁻¹⁶ Wb /m²
Answer : B

Question. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is:
a. qb B/m
b. q (b – a) B/m
c. qa B/m
d. q (b + a) B/2m
Answer : B

Question. A proton of mass 1.67 x 10⁻²⁷ kg and charge 1.6 x 10⁻¹⁹ C is projected with a speed of 2×10⁶ m/ s at an angle of 60° to the X-axis. If a uniform magnetic field of 0.104 Tesla is applied along Y-axis, the path of proton is:
a. A circle of radius = 0.2 m and time period π ×10⁻⁷ s
b. A circle of radius = 0.1 m and time period 2π ×10⁻⁷ s
c. A helix of radius = 0.1 m and time period 2π ×10⁻⁷ s
d. A helix of radius = 0.2 m and time period 4π ×10⁻⁷ s
Answer : B

Question. A coil of 50 turns is situated in a magnetic field b = 0.25weber/m² as shown in figure. A current of 2A is flowing in the coil. Torque acting on the coil will be:

a. 0.15 N
b. 0.3 N
c. 0.45 N
d. 0.6 N
Answer : B

Question. The coil of a galvanometer consists of 100 turns and effective area of 1 square cm. The restoring couple is 10^{– 8} N-m rad. The magnetic field between the pole pieces is 5 T. The current sensitivity of this galvanometer will be:
a. 5 × 10⁴ rad/μ amp
b. 5 × 10^{– 6} per amp
c. 2 × 10^{– 7} per amp
d. 5 rad /μ amp
Answer : D

Question. Consider a magnetic dipole kept in the north south direction. Suppose A₁ , A₂, B₁, B₂ be the four points at the same distance from the dipole towards north, south, east and west of the dipole respectively. The directions of the magnetic field due to the dipole are the same at:
a. A₁ and A₂
b. B₁ and B₂
c. A₁ and B₁
d. A₂ and B₂
Answer : A, B

Question. Which of the following statement (s) is/are correct?
a. Diamagnetism occurs in all materials
b. Diamagnetism is produced due to partial alignment of permanent magnetic dipoles
c. The magnetic field of induced magnetic moment is opposite to the applied field
d. Ferromagnetism is produced due to formation of a large number of small effective regions in the material, called domains and their alignment in external magnetic field
Answer : A, C, D

Question. An infinitely long, straight conductor AB is fixed and a current is passed through it. Another movable straight wire CD of finite length and carrying current is held perpendicular to it and released. Neglect weight of the wire:

a. The rod CD will move upwards parallel to itself
b. The rod CD will move downward parallel to itself
c. The rod CD will move upward and turn clockwise at the same time
d. The rod CD will move upward and turn anti –clockwise at the same time
Answer : C

Question. Three long, straight and parallel wires carrying currents are arranged as shown in the figure. The wire C which carries a current of 5.0 amp is so placed that it experiences no force. The distance of wire C from wire D is then:

a. 9 cm
b. 7 cm
c. 5 cm
d. 3 cm
Answer : A

Question. When a ferromagnetic material goes through a hysteresis the magnetic susceptibility:
a. has a fixed value
b. may be zero
c. may be infinity
d. may be negative
Answer : B, C, D

Question. A proton moving with a constant velocity passes through a region of space without any change in any change in its velocity. If E and B represent the electric and magnetic fields respectively. Then, this region of space may have:
a. E = 0, B = 0
b. E = 0, B ≠ 0
c. E ≠ 0, B = 0
d. E ≠ 0, B ≠ 0
Answer : A, B, D

Question. H⁺ ,He⁺ and O²⁺ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of H, He⁺ and O²⁺ are 1amu, 4amu and 16amu respectively. Then:
a. H⁺ will be deflected most
b. O²⁺ will be deflected most
c. He⁺ and O²⁺ will be deflected equally
d. all will be deflected equally
Answer : A, C

Question. A micro-ammeter has a resistance of 100Ω and a full scale range of 50 μA. It can be used as a voltmeter or as a high range ammeter provided resistance is added to it. Pick the correct range and resistance combination:
a. 50 V range and 10 KΩ resistance in series
b. 10 V range and 20 KΩ resistance in series
c. 5 mA range with 1Ω resistance in parallel
d. 10 mA range with 1Ω resistance in parallel
Answer : B, C

Question. A dip circle is taken to geomagnetic equator. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. Which of the following statements is/are not correct?
a. The needle will stay in a horizontal direction only
b. The needle will stay in vertical direction only
c. The needle will stay in any direction except vertical and horizontal
d. The needle will stay in any direction it is released
Answer : A, B, C

Question. To measure the magnetic moment of a bas magnet, one may use:
a. a tangent galvanometer
b. a deflection galvanometer if the earth’s horizontal field is known
c. an oscillation magnetometer if the earth’s horizontal field is known
d. both deflection and oscillation magnetometer if the earth’s horizontal field is not known
Answer : B, C, D

Question. Which of the following statement is (are) correct in the given figure?
a. net force on the loop is zero
b. net torque on the loop is zero
c. loop will rotate clockwise about axis OO’when seen from O
d. loop will rotate anticlockwise about axis OO’ when seen from O
Answer : A, C

Question. A particle of mass m and charge q, moving with velocity v enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region II is l. Choose the correct choice (s).
a. The particle enters Region III only if its velocity v > qlB / m
b. The particle enters Region III only if its velocity v < qlB / m
c. Path length of the particle in Region II is maximum when velocity
v = qlB / m
d. Time spent in Region II is same for any velocity v as long as the particle returns to Region I
Answer : A, C, D

Question. Two coaxial solenoids 1 and 2 of the same length are set so that one is inside the other. The number of turns per unit length are n₁ and n₂ . The currents i₁ and i₂ are flowing in opposite directions. The magnetic field inside the inner coil is zero. This is possible when:
a. i₁ ≠ i₂ and n₁ = n₂
b. i₁ = i₂ and n₁ ≠ n₂
c. i₁ = i₂ and n₁ = n₂
d. i₁ n₁₂ =  i₂ n₂
Answer : C, D

Question. H⁺ , He⁺ and O⁺⁺ ions having same kinetic energy pass through a region of space filled with uniform magnetic field B directed perpendicular to the velocity of ions. Th e masses of the ions H⁺ ,He⁺ and O⁺⁺ are respectively in the ratio 1: 4 :16. As a result:
a. H⁺ ions will be deflected most
b.O⁺⁺ ions will be deflected least
c. He⁺ and O⁺⁺ ions will suffer same deflection
d. All ions will suffer the same deflection
Answer : A, C

Question. A particle of charge +q and mass m moving under the influence of a uniform electric field Eiˆ and a uniform magnetic field Bkˆ follows trajectory from P to Q as shown in figure. The velocities at P and Q are viˆ and −2vˆj respectively. Which of the following statement(s) is/are correct?

a. E = 3/4 mv² / qa
b. Rate of work done by electric field at P is 3/4 mv³ / a
c. Rate of work done by electric field at P is zero
d. Rate of work done by both the fields at Q is zero
Answer : A, B, D

Comprehension Based

Paragraph –I

Suppose we have a rectangular bar magnet having length, breadth and mass are L, b and w respectively if it is cut in n equal parts along the length as well as perpendicular to the length simultaneously as shown in the figure then

Length of each part L ' = L/√n, breadth of each part b' = b/√n,
Mass of each part w ' = w/n, pole strength of each part m ' = m/√n
Magnetic moment of each part M ' = m' L' = m/√n x  L/√n = M/n
 

Question. A long magnet is cut in two parts in such a way that the ratio of their lengths is 2 : 1. The ratio of pole strengths of both the section is:
a. Equal
b. In the ratio of 2 : 1
c. In the ratio of 1 : 2
d. In the ratio of 4 : 1
Answer : A

Question. A long magnetic needle of length 2L, magnetic moment M and pole strength m units is broken into two pieces at the middle. The magnetic moment and pole strength of each piece will be:
a. ,M/2 , m/2
b. M, m/2
c. M/2 , m
d. M,m
Answer : C

Match the Column

Question. Match the statement of Column with those in Column II:

a. A → 4, B → 3, C → 2, D → 1
b. A → 2, B → 4, C → 3, D → 1
c. A → 1, B → 3, C → 2, D → 4
d. A → 4, B → 1, C → 3, D → 2
Answer : A

Question. Match the statement of Column with those in Column II:

a. A → 4, B → 3, C → 2, D → 1
b. A → 2, B → 4, C → 3, D → 1
c. A → 1, B → 2, C → 3, D → 4
d. A → 4, B → 1, C → 3, D → 2
Answer : C

Question. Column I fives certain situations in which a straight metallic wire of resistance R is used and Colum II gives some resulting effects. Match the statements in Column I with the statements in Column II:

a. A → 1, B → 3,4, C → 4, D → 2,3
b. A → 2, B → 4 C → 4, D → 1,2,3
c. A → 3,4 B → 2 C → 4, D → 1,2,3
d. A → 2, B → 3,4 C → 4, D → 1,2,3
Answer : D

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