# Plus Two Physics Chapter Wise Questions and Answers Chapter 6 Electro Magnetic Induction

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## Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 6 Electro Magnetic Induction

### Plus Two Physics Electro Magnetic Induction NCERT Text Book Questions and Answers

Question 1.
A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0ms-1 at right angles to the horizontal component of the earth’s magnetic field 0.3 × 10-4Wbm2.

1. What is the instantaneous value of the e.m.f? induced in the wire?
2. What is the direction of the e.m.f.?
3. Which end of the wire is at the higher electrical potential?

I – 10m, v = 5.0ms-1, B = 0.30 × 10-4Wbm2
1. Instantaneous value of e.m.f. induced
e = B lv
= 0.30 × 10-4 × 10 × 5.0
or e = 1.5 × 10-3V.

2. The direction of e.m.f. is from eastto west.

3. Since the current is flowing from east to west so the eastern end is at higher potential.

Question 2.
Current in a circuit falls from 5.0Ato 0.0A in 0.1s. If an average e.m.f. of 200V induced, give an estimate of the self-inductance of the circuit.
dl = l2 – l1 = 0.0 – 5.0 = -5.0A,
dt = 0.1s, e = 200V, L=?
Since

or L = 4H

Question 3.
A pair of adjacent coils has a mutual inductance of 1 5H. If the current in one coil changes from 0 to 20A in 0.5s. What is the charge of flux linkage with the other coil?
M= 1.5H, dl = l2 – l1 = 20 – 0 = 20A, dt = 0.5s, dΦ = ?
Since

And also

or dΦ = Mdl
or dΦ = 1.5 × 20 = 30 Wb.

Question 4.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing 25m long if the earth’s magnetic field at the location has a magnitude of 5.0 × 10-4 and the dip angle is 30?
v = 1800 km/h = 500ms-1 I = 25m
Vertical component of B along horizontal
B = -Bsinθ = 5.0 × 10-4 sin30° = 2.5 × 10-4T
e = B/v sinθ
= 500 × 2.5 × 10-4 × 25
= 31 V
The direction ofthe wing is immaterial.

Question 5.
Suppose the loop in Q.6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3T at the rate of 0.02TS-1. If the cut is joined and the loop has a resistance of 1.6Ω, how much power is dissipated by the loops as heat? What is the source of this power?
Induced e.m.f.

= 8 × 2 × 10-4 × 0.02 = 3.2 × 10-5V
Induced current,

= 2 × 10-5A
Power Loss = I2r = (2 × 10-5)2 × 1.6 = 6.4 × 10-10 W. Source of this power is the external agent responsible for changing the magnetic field with time.

### Plus Two Physics Electro Magnetic Induction One Mark Questions and Answers

Question 1.
Eddy current are produced when.
(a) a metal is kept in varying magnetic field
(b) a metal is kept in steady magnetic field
(c) a circular coil is placed in a magnetic field
(d) through a circular coil, current is passed
(a) a metal is kept in varying magnetic field

Question 2.
A100 milli henry coil carries a current of 1 A. energy stored in its magnetic field is
(a) 0.5 J
(b) 1 A
(c) 0.05 J
(d) 0.1 J
(c) 0.05 J
Explanation : E = $$\frac{1}{2}$$ LI2 = $$\frac{1}{2}$$ × (100 × 10-3) × 12
= 0.05 J.

Question 3.
A straight line conductor of length 0.4 m is moved with a speed of 7m/s perpendicular to a magnetic field of intensity 0.9 Wb/m2. The induced e. m. f. across the conductor is.
(a) 5.04 V
(b) 25.2 V
(c) 1.26 V
(d) 2.52
(d) 2.52
Induced e.m.f (s) = B/V
= 0.9 × 0.4 × 7 = 2.52V.

Question 4.
The core of a transformer is laminated because
(a) ratio of voltage in primary and secondary may be increased
(b) energy losses due to eddy currents may be minimized
(c) the weight of the transformer may be reduced
(d) rusting of the core may be prevented
(b)

### Plus Two Physics Electro Magnetic Induction Four Mark Questions and Answers

Question 1.

A conductor XY is moving through a uniform magnetic field of intensity B as shown in figure.

1. Name the emf.
2. The motion of XY towards right side causes an anticlockwise induced current. What will be the direction of magnetic induction in the region A?
3. The length of the conductor XY is 20cm. It is moving with a velocity 50m/s perpendicular to the magnetic fie Id. If the induced emf in the conductor is 2V find the magnitudes of magnetic field?

1. Motional emf/induced emf.

2. Applying right hand grip rule at A direction of magnetic field is away from the paper.

3. ε = Blvsinθ, I = 20cm = 20 × 10-2m, v
= 50m/s, ε = 2v

= 0.2 T.

Question 2.
1. When S1 close bulb glows instantaneously when S2 closes there is a delay in glowing the bulb.(1) This is due to………

• resistance of the coil
• back emf in the coil
• mutual induction
• none

2. Explain the phenomenon self-induction regarding above experiment (2)
3. Draw the graph with energy and current for a inductor. (1)
1. Back emf in the coil.

2. When a.c. current is passed through the coil, a change in flux is produced. This change in flux produces a back e.m.f. in the coil. The phenomena of production of back emf is called self induction.

3. The graph with energy and current for a inductor:

### Plus Two Physics Electro Magnetic Induction Five Mark Questions and Answers

Question 1.
When AC is switched on the thin metallic disc is found to thrown up in air.

1. Which makes the disc to thrown?
2. How will you explain the mechanism behind the movement of disc
3. Write the working principle of induction heater.

1. Eddy current produced in the coil thrown disc into air.
2. Whenever the magnetic flux linked with a metal block changes, induced currents are produced due to this current, disc becomes a magnet. Hence disc thrown up in to air.
3. The change in flux produces eddy current in a metal. The heat produced by eddy current is used for cooking in induction heater.

Question 2.
AC generator is a device based on electromagnetic induction used to convert mechanical energy into electrical energy.

1. Draw a graph showing variation of induced emf over a cycle. Also indicate peak value of emf. (2)
2. How peak value of emf is related to its rms value? (1)
3. A rectangular wire loop of side 4cm × 6cm has 50 turns uniformly from 0.1 Tesla to 0.3 Tesla in 6 × 10-2 second. Calculate induced emf in the coil. (2)

1.

2. Erms = $$\frac{E_{0}}{\sqrt{2}}$$

3.

and Φ = BAN
Φ1 = 0.1 × (24 × 10-4) × 50
= 120 × 10-4wb
Φ2 = 0.3 × (24 × 10-4) × 50
= 360 × 10-4wb

= 40 × 10-2V.