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Learn/ 12th Physics/ Unit 15 /Numericals

Unit 15: Electromagnetic Induction — Numericals

12th Class Physics · Unit 15: Electromagnetic Induction

15.1.A metal rod of length 25 cm is moving at a speed of 0.5 ms⁻¹ in a direction perpendicular to a 0.25 T magnetic field. Find the emf produced in the rod.
Given
Speed of rod v = 0.5 ms-1
Length of rod L = 25 cm = 0.25 m
Magnetic flux density B = 0.25 T
Formula
Induced emf varepsilon = vBL
Working
Substituting values varepsilon = 0.5 × 0.25 × 0.25
= 0.03125 V
Result
Induced emf varepsilon = 3.13 × 10-2 V
15.2.A loop of wire is placed in a uniform magnetic field that is perpendicular to the plane of the loop. The strength of the magnetic field is 0.6 T. The area of the loop begins to shrink at a constant rate of ΔA/Δt = 0.8 m²s⁻¹. What is the magnitude of emf induced in the loop while it is shrinking?
Given
Rate of change of area ΔAΔt = 0.8 m2s-1
Magnetic flux density B = 0.6 T
Number of turns N = 1
Formula
Applying Faraday's law, magnitude of induced emf varepsilon = N Δ PhiΔt
= N vec{B} · frac{Δ vec{A}{Δt}
= N B ΔA cos 0^circΔt
= N B ΔAΔt
Working
Substituting values = 1 × 0.6 × 0.8
= 0.48 JC-1
Result
Induced emf varepsilon = 0.48 V
15.3.An emf of 5.6 V is induced in a coil while the current in a nearby coil is decreased from 100 A to 20 A in 0.02 s. What is the mutual inductance of the two coils? If the secondary has 200 turns, find the change in flux during this interval.
15.4.The current in a coil of 1000 turns is changed from 5 A to zero in 0.2 s. If an average emf of 50 V is induced during this interval, what is the self inductance of the coil? What is the flux through each turn of the coil when a current of 6A is flowing?
Given
Change in current ΔI = 5 A - 0 = 5 A
Time interval Δt = 0.2 s
emf induced varepsilon = 50 V
Steady current I = 6 A
No. of turns of coil N = 1000
Formula
Using formula L = varepsilonΔI / Δt = varepsilon × ΔtΔI
Working
= 50 × 0.25
L = 2 H
Formula
Now, using equation N Phi = LI quad or quad Phi = LIN
Working
Phi = 2 × 61000
= 1.2 × 10-2 Wb
L = 2 H; Phi = 1.2 × 10-2 Wb
15.5.A solenoid coil 10.0 cm long has 40 turns per cm. When the switch is closed, the current rises from zero to its maximum value of 5.0 A in 0.01 s. Find the energy stored in the magnetic field if the area of cross-section of the solenoid is 28 cm².
Given
Length of solenoid l = 10.0 cm = 0.1 m
No. of turns n = 40 per cm = 4000 per m
Area of cross section A = 28 cm2 = 2.8 × 10-3 m2
Stead current I = 5 A
Formula
First, we calculate the inductance L using the equation L = μ0 n2 A l
Working
= (4pi × 10-7) × (4000)2 × 2.8 × 10-3 × 0.1
= 5.63 × 10-3 H
Formula
Energy stored Um = 12 L I2
Working
= 12 (5.63 × 10-3) × (5)2
= 7.04 × 10-2 J
Um = 7.04 × 10-2 J
15.6.An alternating current generator operating at 50 Hz has a coil of 200 turns. The coil has an area of 120 cm². What should be the magnetic field in which the coil rotates in order to produce an emf of maximum value of 240 volts?
Given
Frequency of rotation f = 50 Hz
No. of turns of the coil N = 200
Area of the coil A = 120 cm2 = 1.2 × 10-2 m2
Maximum emf varepsilonmax = 240 V
Formula
Using omega = 2pi f
= 2 × 227 × 50 = 314.3 rad s-1
Using varepsilon0 = N omega A B
B = varepsilon0N omega A
Working
B = frac{240}{200 × 314.3 × 1.2 × 10-2
= 0.32 T
B = 0.32 T
15.7.A permanent magnet D.C motor is run by a battery of 24 volts. The coil of the motor has a resistance of 2 ohms. It develops a back emf of 22.5 volts when driving the load at normal speed. What is the current when motor just starts up? Also find the current when motor is running at normal speed.
Given
Operation voltage V = 24 V
Resistance of the coil R = 2 Ω
Back emf varepsilon = 22.5 V
Formula
(i) When motor just starts up, the back emf ε = 0 V = varepsilon + IR
Working
24 = 0 + I × 2
I = 242 = 12 A
Formula
(ii) When motor runs at normal speed, ε = 22.5 V then using V = varepsilon + IR
Working
V = 22.5 + I × 2
I = 24 - 22.52
= 0.75 A
I1 = 12 A; I2 = 0.75 A
15.8.The turns ratio of a step up transformer is 50. A current of 20 A is passed through its primary coil at 220 volts. Obtain the value of the voltage and current in the secondary coil assuming the transformer to be ideal one.
Given
Turn ratio NsNp = 50
Current from primary Ip = 20 A
Primary voltage Vp = 220 V
Formula
Using the equation VsVp = NsNp
Working
Vs220 = 50
Vs = 50 × 220 = 11000 volt
Formula
Since Is = VpVs × Ip
Working
= 150 × 20
= 0.4 A
Vs = 11000 volt; Is = 0.4 A