CBSE Class 12 Physics Electro Magnetic Induction And Alternating Currents Notes

4. 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 ΦB = B.A = BACosθ where θ is the angle between B and A.

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.

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 produce a current, which opposes 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  = 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.

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 self induced e.m.f is given by

di e L dt

 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.

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 = μ0n2Al where A is the area of cross-section 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 n1n2Al 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.

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

                                           E= -dΦ/dt = NBAωSinωt                                               E = E0Sinωt

                                            * Current induced in the coil I = E/R = E₀Sinωt/R = I₀Sinωt

17. An alternating voltage E = E ₀Sinωt, applied to a resistor R drives a current I = I₀Sinωt in the resistor, I₀ = E ₀/R where E₀ & I₀ are the peak values of voltage and current. (also represented by V_m & I_m)

18. The root mean square value of a.c. may be defined as that value of steady current which would generate the same amount of heat in a given resistance in a given time as is done by the a.c. when passed through the same resistance during the same time.

                                             I_rms = I₀/√2 = 0.707i₀

                                             Similarly, v_rms = v₀/√2 = 0.707v_0.

19. For an a.c. E = E_m Sin ωt applied to a resistor, current and voltage are in phase.

20. In case of an a.c. circuit having pure inductance current lags behind e.m.f by a phase angle 90°.

                                          E = E_m Sin ωt and i = i_m Sin (ωt-Π/2)

                                          I_m = E_m/X_L; X_L = ωL is called inductive reactance.

21. In case of an a.c. circuit having pure capacitance, current leads e.m.f by a phase angle of 90°.

                                         E = E_mSinωt and I= I_mSin(ωt+π/2) where

                                         I_m = E_m/XC and XC = 1/ωC is called capacitive reactance.

22. In case of an a.c. circuit having R, L and C, the total or effective resistance of the circuit is called impedance (Z).

23. Average power loss over a complete cycle in an LCR circuit is

                                         P = E_rmsI_rmsCosΦ

                                        * In a purely resistive circuit Φ = 0; P = V_RMSI_RMS.

                                       * In a purely inductive circuit Φ = Π/2; P = 0.

                                       * In a purely capacitive circuit Φ = Π/2; P = 0.

24. In an LCR circuit, the circuit admits maximum current if X_C = X_L, so that Z = R and resonant

25. Q factor of series resonant circuit is defined as the ratio of voltage developed across the inductance or capacitance at resonance to the applied voltage across ‘R’,

                                     Q = ω₀L/R or 1/ ω₀CR.

                                    In an ideal transformer, EPIP = ESIS.

                                    If N_S>N_P; E_S>E_P & I_S<I_P - step up.

                                    If N_P>N_S; E_P>E_S & I_P<I_S – step down

27. A circuit containing an inductor L and a capacitor C (initially charged) with no a.c. source and no resistors exhibits free oscillations of energy between the capacitor and inductor. The charge q satisfies the equatio

Question. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque required to maintain the needle in this position will be:
a. 3 W
b. −W
c. (√3/2) W
d. 2W
Answer : A

Question. An iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now it’s magnetic moment will be:
a. M
b. 2M/π
c. M/π
d. Mπ
Answer : B

Question. A short bar magnet with its north pole facing north forms a neutral point at P in the horizontal plane. It the magnet is rotated by 90° in the horizontal plane, the net magnetic induction at P is: (Horizontal component of earth's magnetic field = B_H)
a. 0
b. 2 B_H
c. (√5/2) B_H
d.√5 B_H
Answer : D
 

Question. A bar magnet with its poles 25cm apart and of pole strength 24 amp × m rests with its centre on a frictionless pivot. A force F is applied on the magnet at a distance of 12 cm from the pivot so that it is held in equilibrium at an angle of 30° with respect to a magnetic field of induction 0.25 T. The value of force F is:
a. 5.62 N
b. 2.56 N
c. 6.52 N
d. 6.25 N
Answer : D

Question. Two identical bar magnets with a length 10 cm and weight 50 gm – weight are arranged freely with their like poles facing in a arranged vertical glass tube. The upper magnet hangs in the air above the lower one so that the distance between the nearest pole of the magnet is 3mm. Pole strength of the poles of each magnet will be:

a. 6.64 amp × m
b. 2 amp × m
c. 10.25 amp × m
d. None of these
Answer : A

Question. If the angles of dip at two places are 30° and 45° respectively. Then the ratio of horizontal components ofearth's magnetic field at the two places will be:
a. √3 : √2
b. 1 : √2
c. 1 : 3
d. 1 : 2
Answer : A

Question. Earth's magnetic field may be supposed to be due to a small bar magnet located at the centre of the earth. If the magnetic field at a point on the magnetic equator is 0.3 x 10⁻²⁴ T Magnet moment of bar magnet is:

a. 7.8×10⁸ amp×m²
b. 7.8×10²² amp×m²
c. 6.4×10²² amp×m²
d. None of these
Answer : B

Question. A magnet of magnetic moment 20 C.G.S. units is freely suspended in a uniform magnetic field of intensity 0.3 C.G.S. units. The amount of work done in deflecting it by an angle of 30° in C.G.S. units is:
a. 6
b. 3√3
c. 3(2 − √3)
d. 3
Answer : C

Question. The magnetic field at a point X on the axis of a small bar magnet is equal to the field at a point Y on the equator of the same magnet. The ratio of the distance of X and Y from the centre of the magnet is:
a. 2⁻³
b. 2^{−1/ 3}
c. 2³
d. 2^{1/ 3} 
Answer : D

Question. When 2 amp current is passed through a tangent galvanometer, it gives a deflection of 30°. For 60° deflection, the current must be:
a. 1 amp.
b. 2√3 amp.
c. 4 amp.
d. 6 amp.
Answer : D

Question. In vibration magnetometer the time period of suspended bar magnet can be reduced by:
a. Moving it towards South Pole
b. Moving it towards North Pole
c. Moving it toward equator
d. Anyone them
Answer : C

Question. A certain amount of current when flowing in a properly set tangent galvanometer produces a deflection of 45°. If the current be reduced by a factor of √3, the deflection would:
a. Decrease by 30°
b. Decreases by 15°
c. Increase by 15°
d. Increase by 30°
Answer : B

Question. The angle of dip at a place is 60°. A magnetic needle oscillates in a horizontal plane at this place with period T.
The same needle will oscillate in a vertical plane coinciding with the magnetic meridian with a period:
a. T
b. 2T
c. T/2
d. T/√2
Answer : D

Question. The coereivity of a small bar magnet is 4 × 103 Amp/m. It is inserted inside a solenoid of 500 turns and length 1 m to demagnetise it. The amount of current to be passed through the solenoid will be:
a. 2.5 A
b. 5 A
c. 8 A
d. 10 A
Answer : C

Question. Two magnets are held together in a vibration magnetometer and are allowed to oscillate in the earth's magnetic field. With like poles together 12 oscillations per minute are made but for unlike poles together only 4 oscillations per minute are executed. The ratio of their magnetic moments is:
a. 3 : 1
b. 1 : 3
c. 3 : 5
d. 5 : 4
Answer : D

Question. The magnetic moment of a current carrying loop is 2.1 10^-25 amp x m² . The magnetic field at a point on its axis at a distance of 1 Å is:
a. 4.2 x 10^-2 weber /m²
b. 4.2 x 10^-3 weber /m²
c. 4.2 x 10^-4 weber /m²
d. 4.2 x 10^-5 weber /m²
Answer : A

Question. Find the position of point from wire 'B' where net magnetic field is zero due to following current distribution:

a. 4 cm
b. (30/7)cm
c. (12/7)cm
d. 2 cm
Answer : C

Question. Find out the magnitude of the magnetic field at point P due to following current distribution:

Answer : A

Question. A magnet makes 40 oscillations per minute at a place having magnetic field intensity B_H = 0.1 x 10⁻⁵ At another place, it takes 2.5 sec to complete one-vibration. The value of earth's horizontal field at that place:
a. 0.25 10⁻⁶ T
b. 0.36 10⁻⁶ T
c. 0.66 10⁻⁸ T
d. 1.2 10⁻⁶ T
Answer : B

Question. Two parallel, long wires carry currents i₁ and i₂ with i₁ / > i₂ when the currents are in the same direction, the magnetic field at a point midway between the wires is 10 μT. If the direction of i₂ is reversed, the field becomes 30μT. The ratio i₁ / i₂ is:
a. 4
b. 3
c. 2
d. 1
Answer : C

Question. Two infinite length wires carries currents 8A and 6A respectively and placed along X and Y-axis. Magnetic field at a point P(0, 0, d)m will be:

Answer : D

Question. An equilateral triangle of side 'a' carries a current i then find out the magnetic field at point P which is vertex of triangle:

Answer : B

Question. Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. The value of magnetic field at the centre of the loop assuming uniform wire is:

Answer : B

Question. The average radius of a toroid made on a ring of nonmagnetic material is 0.1 m and it has 500 turns. If it carries 0.5 ampere current, then the magnetic field produced along its circular axis inside the toroid will be:
a. 25 x 10⁻² Tesla
b. 5 x 10⁻²Tesla
c. 25 x 10⁻⁴ Tesla
d. 5 x 10⁻⁴ Tesla
Answer : D

Question. The units for molar susceptibility:
a. m³
b. kg-m^–3
c. kg^–1 m³
d. No units
Answer : A

Question. A straight section PQ of a circuit lies along the X-axis from x = - a/2 to x = a/2 x = and carries a steady current i. The magnetic field due to the section PQ at a point X = + a will be:
a. Proportional to a
b. Proportional to a²
c. Proportional to 1/a
d. Zero
Answer : D

Question. A wire in the form of a square of side a carries a current i. Then the magnetic induction at the centre of the square wire is: (Magnetic permeability of free space = μ0)

a. μ₀i / 2πa
b. μ₀i√2 / πa
c. 2√2μ₀i / πa
d. μ₀i / √2πa
Answer : C

Integer

Question. A current 1 amp is flowing in the sides of an equilateral triangle of side 4.5×10^–2m. The magnetic field at the centorid of the triangle in the units of (10^–5) is:
Answer : 4

Question. A length of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. The same length is now best more sharply to give double loop of smaller radius. If the same current I is passed, the ratio of the magnitude of magnetic field at the centre with its first value is:
Answer : 4

Question. A galvanometer has a sensitivity of 60 divisions / ampere. When a shunt is used its sensitivity becomes 10 divisions / ampere. Wheat is the value of shunt (in ohm) used if the resistance of the galvanometer is 20Ω
Answer : 4

Question. There are two infinite long parallel straight current carrying wires, A and B separated by a distance r (Fig.). The current in each wire is I. The ratio of magnitude of magnetic field at points P and Q when points P and Q lie in the plane of wires is:

Answer : 8

Question. The coercivity of a bar magnet is 120 A/m. it is to be demagnetized by placing it inside a solenoid of length 120 cm and number of turns 72. The current (in A) flowing through the solenoid is:
Answer : 2

Question. A length of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give double loop of smaller radius. If the same current I is passed, the ratio of the magnitude of magnetic field at the centre with its first value is:

Answer : 4
 

Please refer to attached file for Electro Magnetic Induction And Alternating Currents.