The phenomenon of producing an electric current in a coil of wire with the help of a magnetic field is called electromagnetic induction. Many devices such as transformers and generators work on the principle of electromagnetic induction. The laws of Physics that govern the phenomenon of electromagnetic induction are called Faraday’s Laws of Electromagnetic Induction.
The phenomenon in which an emf is induced in a coil due to the change of current through the coil itself is known as self-induction. If the change of current in one coil induces an emf in another neighbouring coil, it is called mutual induction.
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Electromagnetic Induction and Magnetism JEE Advanced Questions
Question 1) A step-down transformer has 50 turns on secondary and 1000 turns on the primary winding. If a transformer is connected to 220 V and 1 A AC source, what is the output current of the transformer?
1) 1/20 A
2) 20 A
3) 100 A
4) 2 A
Answer: 2) 20 A
Solution:
Number of turns in the primary winding, Np= 1000 turns
Number of turns in the secondary winding, Ns= 50 turns
Current in primary windings, Ip = 1 A
Current in the secondary windings, Is = (Np/Ns)Ip
= (1000/50) x 1
= 20 A
Question 2) The time constant of an LR circuit is 20 ms. The circuit is connected at t = 0 and the steady-state current is found to be 4A. Find the current at 80 ms.
1) 0.98 A
2) 1 A
3) 0.44 A
4) 0.88 A
Answer: 4) 0.88 A
Solution:
I = I0(1 – et/Τ)
at t = 0, I = I0
At t = 0, I = 4 A hence I0 = 4 A
I = 4(1 – et/Τ)
Given Τ = 80 ms = 80 × 10-3 s
Current at 20 ms = 20 × 10-3 s
I = 4[1 – e(-20 × 10-3 /80 × 10-3)]
= 4(1 – e-(1/4))
= 4(1 – e-0.25)
= 4(1 – 0.778)
I = 0.88 A
Question 3) A coil of inductance 300 mH and resistance 2Ω is connected to a source of voltage 2V. The current reaches half of its steady-state value in………
1) 0.15 sec
2) 0.3 sec
3) 0.05 sec
4) 0.1sec
Answer: 4) 0.1sec
Solution:
I = I0(1 – e-Rt/L)
I0/2 = I0(1 – e-Rt/L)
e-Rt/L = (1/2)
Taking log on both sides
Rt/L = ln2
⇒ t = (L/R)ln2
t = (300 × 10-3/2) × 0.693
= 0.1 sec
Question 4) The same current is flowing in two alternating circuits. The first circuit contains only inductance and the other contains only a capacitor. If the frequency of the emf of ac is increased the effect on the value of the current will be.
1) Increase in the first circuit and decrease in other
2) Increase in both the circuit
3) Decrease in both the circuit
4) Decrease in the first and increase in other
Answer: 4) Decrease in the first and increase in other
Solution:
For the first current, i = V/Z
An increase in ω will cause a decrease in i
For the second circuit,
An increase in ω will cause an increase in i.
Question 5) A 50 turns circular coil has a radius of 3 cm, it is kept in a magnetic field acting normal to the area of the coil. The magnetic field B increased from 0.10 tesla to 0.35 tesla in 2 milliseconds. The average induced e.m.f. in the coils is
1) 1.77 volts
2) 17.7 volts
3) 177 volts
4) 0.177 volts
Answer: 2) 17.7 volts
Solution:
N = 50 turns
r = 3 × 10-2 m
B1 = 0.10
B2 = 0.35
t = 2 × 10-3 sec
Area = πr2
average induced emf = (Φ1 – Φ2)/t
Φ1 = NB1A
Φ2 = NB2A
e = (NB1A – NB2A)/t
e = (NA/t) [B2 – B1]
e = 17.675 volts
= 17.7 volts
Question 6) The rms value of ac of 50 Hz is 10 amp. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be,
1) 2 × 10-2 sec and 14.14 amp
2) 1 × 10-2 sec and 7.07 amp
3) 5 × 10-3 sec and 7.07 amp
4) 5 × 10-3 sec and 14.14 amp
Answer: 4) 5 x 10-3 sec and 14.14 amp
Solution:
Time for reaching the maximum of peak value from 0
T = 1/f
T = 1/4f
= 1/(4 × 50)
= 1/200
= 5 × 10-3 sec
I0 = √2Irms
= 10 × √2
= 14.14 A
Question 7) In a circular coil. when no. of turns is doubled & resistance becomes half of the initial, then inductance becomes_____.
1) 4 times
2) 2 times
3) 8 times
4) No change
Answer: 1) 4 times
Solution:
The formula for self-inductance L = μN2A/l
So, L ∝ N2
L does not depend on the resistance
So N will become 2N
L will be equal to 4(μN2A/l)
= 4 times of the initial
Question 8) A rectangular coil of 20 turns and area of cross-section 25 sq cm has a resistance of 100 ohm. If a magnetic field which is perpendicular to the plane of the coil changes at the rate of 1000 tesla per second, the current in the coil is
1) 1.0 ampere
2) 50 ampere
3) 0.5 ampere
4) 5.0 ampere
Answer: 3) 0.5 ampere
Solution:
i = ε/R
= 0.5 A
Question 9) A solenoid is 1.5 m long, and its inner diameter is 4.0 cm. It has three layers of windings of 1000 turns each and carries a current of 2.0 amperes. The magnetic flux for a cross-section of the solenoid is nearly
1) 2.5 × 10–7 weber
2) 6.31 × 10–6 weber
3) 5.2 × 10–5 weber
4) 4.1 × 10–5 weber
Answer: 2) 6.31 × 10–6 weber
Solution:
L = 1.5 m
I = 2 A
Total number of turns = 3 x 1000 = 3000
Turns per unit meter = n = 3000/1.5 = 2000
Radius, r = 4/2 = 2 cm = 2 × 10-2 m
Φ = BAcosθ
B = μ0nl
A = πr2
Φ = (μ0nl) × (πr2) × cos 0
= 4π × 10-7 × 2000 × 2 × π × (2 × 10-2)2
= 6.31 × 10-6 weber
Also Read:
Electromagnetic Induction JEE Main Previous Year Questions With Solutions
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