Electromagnetic Induction Test

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Electromagnetic Induction Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

The impedance of an AC circuit, which contains a capacitive reactance of 5 ohms and inductive reactance of 8 ohms, will be

A
1.6 ohms

B
40 ohms

C
3 ohms

D
13 ohms

Solution

X = X_L - X_c
X = 8 -5 - 3 Ώ
Question 2
1 / -0

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm²placed inside the solenoid normal to its axis. Calculate the induced emf in the loop, if the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s.

A
7.54 × 10⁻⁶ V

B
0.754 × 10⁻⁶ V

C
7.54 × 10⁻⁷ V

D
2.4 × 10⁻⁶V

Solution

Number of turns of the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm² = 2 × 10⁻⁴ m²_.Current carried by the solenoid changes from 2 A to 4 A. Change in current in the solenoid, di = 4 − 2 = 2 A. Change in time, dt = 0.1 s. Induced emf in the solenoid is given by Faraday's law as:
e = dΦ / dt
where, Φ = BA (B = μ0ni)
μ0 = Permeability of free space = 4π×10⁻⁷H/m
e = d BA/dt = A μ0 n di/dt = 2×10−4 × 4π×10⁻⁷ × 1500 × 2/0.1 = 7.54 ×10⁻⁶V
Question 3
1 / -0

A circular coil of radius 10 cm, 500 turns and resistance 2Ω is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Determine the magnitudes of the current induced in the coil, where the magnetic field at the place is 3.0 × 10^–5 T.

A
0.94 × 10^-³A

B
1.9 × 10^–3 A

C
0.471 × 10^–3 A

D
0.3 × 10^–3 A

Solution

Initial flux through the coil, Φ = BA cosθ = 3.0 × 10–5 × (π ×10–2) × cos 0º = 3π × 10–7 Wb. Final flux after the rotation, Φ= 3.0 × 10–5× (π×10–2) × cos 180° = –3π× 10–7 Wb. Therefore, estimated value of the induced emf is, ε= N ΔΦ / Δt= 500 × (6π× 10–7)/0.25 = 3.8 × 10–3VI = ε / R= 1.9 × 10–3 A
Question 4
1 / -0

A 20-turn square coil of side 8.0 mm is pivoted at the centre and placed in a magnetic field of flux density 0.010 T, such that two sides of the coil are parallel to the field and two sides are perpendicular to the field. A current of 5.0 mA is passed through the coil. What is the magnitude of the torque acting on the square coil?

A
0.1 x 10⁴ N m

B
0.1 x 10^-8 N m

C
6.4 x 10^-8 N m

D
62.5 x 10⁸ N m

Solution

Magnetic force on asingle coil = BIL
0.01 X 5.0 X 10^-3X 8 X 10^-3
4.0 X 10^-7
Torque acting on the coil = 20 x 4.0 x 10^-7 x 8.0 x 10^-3= 6.4 x 10^-⁸ N m
Question 5
1 / -0

A plane of wingspan 30 m flies through a vertical field of strength 5 x 10^-4T. Calculate the emf induced across wing tips if its velocity = 150 ms^-1.

A
0.0001 V

B
2.25 V

C
0.0025 V

D
0.09

Solution

Area = 30 x 150 = 4500 m². Flux = ɸ= BA = 5 x 10-4 x 4500 = 2.25 Wb. So induced emf is E = Δɸ / Δt = 2.25 / 1= 2.25 V.
Question 6
1 / -0

A wire 88 cm long bent into a circular loop is kept with plane of the coil perpendicular to the magnetic induction 2.5 Wb/m². With in 2 s the soil is changed to a square and magnetic induction is increased by 0.5 Wb/m². Calculate the e.m.f. induced in the wire.

A
0.10334 V

B
4.4 mV

C
0.0154 V

D
0.4774 V

Solution

L = 0.88 m; B1= 2.5 Wb/m²; dt = 2 s; B2= 3 Wb/m²; e = ?
ø1= A B1=πr²B1
; Here, 2πr = L = 0.88
∴r = 0.88 / 2π
∴ø1=πx 0.88 x 0.88 x 2.5 / 4 π²
= 0.22 x 0.88 x 2.5 / 3.14 = 0.1540
ø2= A B2= L²B2
; Here L = 0.88 / 4 = 0.22
∴ø2= 0.22 x 0.22 x 3 = 0.1452
∴dø = 0.1540–0.1452 = 0.0088
∴e = dø / dt = 0.0088 / 2 = 0.0044 V = 4.4 mV

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Electromagnetic Induction Test - 3

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Electromagnetic Induction Test - 3

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Question 1

1.6 ohms

40 ohms

3 ohms

13 ohms

Solution

Question 2

7.54 × 10−6 V

0.754 × 10−6 V

7.54 × 10−7 V

2.4 × 10−6 V

Solution

Question 3

0.94 × 10-3 A

1.9 × 10–3 A

0.471 × 10–3 A

0.3 × 10–3 A

Solution

Question 4

0.1 x 104 N m

0.1 x 10-8 N m

6.4 x 10-8 N m

62.5 x 108 N m

Solution

Question 5

0.0001 V

2.25 V

0.0025 V

0.09

Solution

Question 6

0.10334 V

4.4 mV

0.0154 V

0.4774 V

Solution

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7.54 × 10⁻⁶ V

0.754 × 10⁻⁶ V

7.54 × 10⁻⁷ V

2.4 × 10⁻⁶V

0.94 × 10^-³A

1.9 × 10^–3 A

0.471 × 10^–3 A

0.3 × 10^–3 A

0.1 x 10⁴ N m

0.1 x 10^-8 N m

6.4 x 10^-8 N m

62.5 x 10⁸ N m