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Electromagnetic Induction and Alternating Current

Description: Electromagnetic Induction and Alternating current
Number of Questions: 15
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Tags: Electromagnetic Induction and Alternating current Electromagnetic Induction
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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?

  1. 0.1 x 104 N m

  2. 0.1 x 10-8 N m

  3. 6.4 x 10-8 N m

  4. 62.5 x 108 N m

  5. 0.025 x 10-4 N m


Correct Option: C
Explanation:

Magnetic force on asingle coil = BIL 0.01 X 5.0 X 10-3 X 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

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

  1. 0.0001 V

  2. 2.25 V

  3. 0.0025 V

  4. 0.09

  5. 0.44 V


Correct Option: B
Explanation:

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.

A field of 5 x 10-2 Wb/m2 acts at right angles to the coil of area 0.2 m2 having 100 turns. The coil is removed from the field in 0.05 second. Find the emf induced in the coil in volt?

  1. 25 volt

  2. 1 volt

  3. 40 volt

  4. 20 volt

  5. 8000 volt


Correct Option: D
Explanation:

The magnitude of the induced emf ε = dΦB / dt = - d NBA / dt = - N B dA / dt = - 100 x 5 x 10-2 x 0.2 / 0.05 = 20 volt.

The magnetic flux through a loop of resistance 0.1Ω is varying according to the relation ø = 6 t 2 + 7 t + 1 where ø is in milli Wb and t is in seconds. What is the e.m.f. induced in the loop at t = 1 s and the magnitude of the current?

  1. 9 mV, 190 mA

  2. 130mV, 130 mA

  3. 13mV, 1.3 mA

  4. 19mV, 5.2 A

  5. 19mV, 1.9 mA


Correct Option: A
Explanation:

R = 0.1 Ω; ø = 6 t2+ 7 t + 1 ; t = 1 s; e = ? I = ? e = dø/dt = 6 x 2 t + 7 = 12 t + 7 ∴At t = 1, e = 12 + 7 = 19 mV Now I = e / R= 19 x 10-3/ 0.1 = 19 x 10-2 = 190 mA.

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed 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.

  1. 7.54 × 10−6 V

  2. 0.754 × 10−6 V

  3. 7.54 × 10−7 V

  4. 2.4 × 10−6 V

  5. 3.768 × 10−7 V


Correct Option: A
Explanation:

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 m2Current 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×10−7 H/m e = d BA/dt = A μ0 n di/dt = 2×10−4 × 4π×10−7 × 1500 × 2/0.1 = 7.54 ×10−6 V

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.

  1. 0.10334 V

  2. 4.4 mV

  3. 0.0154 V

  4. 0.4774 V

  5. 0.2349 V


Correct Option: B
Explanation:

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

A square wire loop of 4 cm sides is perpendicular to a magnetic field of 5 mT. Calculate the magnetic flux through the loop.

  1. 20 x 10-5 Wb

  2. 8 x 10-6 Wb

  3. 0.125 Wb

  4. 3.125 Wb

  5. 0.32 Wb


Correct Option: B
Explanation:

Area of the loop, A = 4 cm x 4 cm = 16 x 10- 4 m2 , Magnetic field, B = 5 mT = 5 x 10- 3 T, Angle, θ = 00, Magnetic flux through the loop, Φ = BA cos θ = 5 x 10- 3 x 16 x 10- 4 cos 0= 8 x 10- 6 Wb

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.

  1. 0.94 × 10-3 A

  2. 1.9 × 10–3 A

  3. 0.471 × 10–3 A

  4. 0.3 × 10–3 A

  5. 0.6 × 10–3 A


Correct Option: B
Explanation:

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

The vertical component of the earth's magnetic field in a place is 3 x 10-5 T. What is the potential difference between the ends of the axle of a car, which are 1.5 m apart, when car's velocity is 20 m/s?

  1. 0.1 x 10-5 V

  2. 9.0 x 10-4 V

  3. 0.025 x 105 V

  4. 40 x 10-5 V

  5. 0.23 x 10-5 V


Correct Option: B
Explanation:

The potential difference between the ends of the axle of a car is given by ε = Bvvl = 3.0 x 10- 5 x 20 x 1.5 = 9.0 x 10- 4 V.

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

  1. 0.0375 Wb

  2. 26.66 Wb

  3. 13.33 Wb

  4. 30 Wb

  5. 0.075 Wb


Correct Option: D
Explanation:

Δ (NΦ) = MΔ I1 = 1.5 x (20 – 0) = 30 Wb

An aeroplane with a wing span of 50 m is flying horizontally with a speed is 200 m/s. The horizontal and 9 vertical components of Earth's magnetic field are 0.3 gauss and 0.5 gauss respectively. Find the potential difference between the wing tips of the aeroplane induced by its motion through the Earth's magnetic field.

  1. 0.5 x 104 V

  2. 2 V

  3. 0.3 V

  4. 0.5 V

  5. 25 x 104 V


Correct Option: D
Explanation:

Wing span l = 50 m. Speed v = 200 m/sHorizontal component of earth's field BH = 0.3 gauss = 0.3 x 10- 4 T. Vertical component of earth's field BV = 0.5 gauss = 0.5 x 10- 4 T. An aeroplane flying horizontally cuts only the vertical component of the earth's field. So, the potential difference between the wing tips of the aeroplane is ε = Bvvl = 0.5 x 10- 4 x 200 x 50 = 0.5 V

A circular coil of 100 turns and an area of 10 cm2 rotates about a diameter with an angular velocity 100 rad/s in a uniform magnetic field of 0.10 T. The axis of rotation is perpendicular to the field. The coil has a resistance of 10Ω. The emf induced in the coil is supplied to an external resistance of 10 ohm. Calculate the peak current flowing through the external resistance.

  1. 20 A

  2. 0.05 A

  3. 5 A

  4. 0.2 A

  5. 2000 A


Correct Option: B
Explanation:

Peak emf is given byε0 = NBAω N = 100, B = 0.10 T, A = 10 cm2 = 10 x 10- 4 m2, ω = 100 rad/s So, ε0 = NBAω = 100 x 0.10 x 10 x 10- 4 x 100 = 1 V . Peak current through the external resistance = ε0 / (10 + 10) = 1/20 = 0.05 A.

The rails of a railway track separated by 1 m are connected to a millivoltmeter. The rails are insulated from each other and the ground. A superfast train travels at a speed of 135 km/h. What will the reading of the millivoltmeter if the horizontal component of the Earth's magnetic field at that place is 0.2 Gauss?

  1. 5.33 mv

  2. 0.75 mv

  3. 1875 mv

  4. 6750 mv

  5. 2.7 mv


Correct Option: B
Explanation:

The reading of the millivoltmeter = motional emf produced across the axle = BvlSpeed v = 135 km/h = 135 x 5/18 m/s = 37.5 m/sMagnetic field B = 0.2 G = 0.2 x 10- 4 T Length l = 1 m Therefore, the reading of the millivoltmeter = 0.2 x 10- 4 x 37.5 x 1 = 0.00075 V = 0.75 mV

A square loop of side 10 cm and resistance 1 ohm is moved towards right with a constant velocity v. The left arm of the loop is in a uniform magnetic field of 2T. The field is perpendicular to the plane and is coming out of it. The loop is connected to a network of resistors each of value 3 ohm. With what speed should the loop be moved so that a steady current of 1 mA flows in the loop?

  1. 50 m/sec

  2. 5 m/sec

  3. 2 cm/sec

  4. 80 cm/sec

  5. 12.5 m/sec


Correct Option: C
Explanation:

The network of resistors forms a balanced Wheatstone bridge,  and hence resistor of 3 ohm in the middle can be neglected. Remaining resistors have an equivalent resistance = 3 ohm. Total resistance = 3 + 1 = 4 ohm. Induced emf, Bvl = IR Or 2 x v x 0.10 = 1 x 10- 3 x 4. This gives v = 2 cm/sec.

A rod of length 1 m aligned in east-west direction falls vertically downward at a constant speed of 72 km/h at a place where the vertical component of the Earth's magnetic field is 1/2 √3 gauss and angle of dip 300. Find the potential difference between two ends of the rod.

  1. 0.0069 x 10-4 V

  2. 10-3 V

  3. 36 x 10-4 V

  4. 40 x 104 V

  5. 0.025 x 10-4 V


Correct Option: B
Explanation:

Length of rod l = 1 m,  Speed v = 72 km/h = 20 m/s, Angle of dip δ = 300. A rod falling vertically along east-west direction cuts only the horizontal component of the earth's field. Vertical component of earth's field Bv = 1/2 √3 gauss = (1/2 √3) x 10- 4 T. Horizontal component of earth's field BH = Bv / tan δ = (1/2 √3)/(1 / √3) gauss = 0.5 x 10-4 T.  So, the potential difference between the ends of is ε = BHvl = 0.5 x 10- 4 x 20 x 1 = 10- 3 V

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