Electromagnetic induction | Class 12 NEB Physics | Numerical problem solution

The magnetic flux passing perpendicular to the plane of coil is given by φ = 4t²+5t+2 where φ is in weber and t is in second. Calculate the magnitude of instantaneous emf induced in the coil when t = 2 sec.
Solution:
Given,
magnetic flux, φ = 4t²+5t+2
time, t = 2 sec
induced emf, ε = ?

From Faraday's law, \[\begin{align*} \epsilon& = -\frac{d\phi}{dt}\\ &=-\frac{d(4t^2+5t+2)}{dt}\\ &=-(8t+5)\\ \end{align*}\] At t = 2 sec, the induced emf is , \[\begin{align*} \epsilon&=-(8 \times 2 + 5)\\ \therefore \epsilon& = -21\hspace{0.1cm}\text{V}\\ \end{align*}\]
Thus, the magnitude of induced emf is 21 V.

A straight conductor of length 25 cm is moving perpendicular to its length with a uniform speed of 10 m/s making an angle of 45⁰ with a uniform magnetic field of 10 T. Calcuate the emf induced across its length.
Solution:
Given,
length, l = 25 cm = 0.25 m
speed, v = 10 m/s
magnetic field, B = 10 T
angle between the velocity vector of motion of conductor and magnetic field, θ = 45⁰
emf induced, ε = ?

Now, the emf induced in a straight current carrying conductor is, \[\begin{align*} \epsilon&=Blvsin\theta\\ &=10 \times 0.25 \times 10 \times sin45^\circ\\ &=17.67 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the emf induced across the length is 17.67 V .

A coil of 100 turns, each of area 2×10^-3 m² has a resistance of 12Ω. It lies in a horizontal plane in a vertical magnetic flux density of 3×10^-3 Wbm^-2. What charge circulates through the coil if its ends are short-circuited and the coil is rotated through 180⁰ about a diametrical axis?
Solution:
Given,
number of turns of coil, N = 100 turns
area of each coil, A = 2×10^-3 m²
resistance of coil, R = 12Ω
magnetic flux density, B = 3×10^-3 Wbm^-2
charge circulated through the coil, \(\Delta q\) = ?

We know, the emf induced in the coil can be written as, \[\begin{align*} |{\epsilon}|&=IR\\ \frac{\Delta\phi}{\Delta t}&=\frac{\Delta q}{\Delta t} R\\ \Delta q&=\frac{\Delta \phi}{R}\\ \end{align*}\]
The change in flux is, \[\begin{align*} \Delta \phi&=NBAcos0^\circ - NBA cos180^\circ\\ &=NBA+NBA\\ &=2NBA\\ &=2 \times 100 \times 3\times 10^{-3} \times 2 \times 10^{-3}\\ \Delta \phi&=1.2\times 10^{-3} \hspace{0.1cm} \text{wbm}^2\\ \end{align*}\]
Now, the charge circulated through the coil is, \[\begin{align*} \Delta q&=\frac{\Delta \phi}{R}\\ &=\frac{1.2 \times 10^{-3}}{12}\\ \therefore \Delta q&= 10^{-4} \hspace{0.1cm} \text{C}\\ \end{align*}\]
Thus, the charge circulated through the coil is 10^-4 C.

An aircraft with a wingspan of 40 m flies with a speed of 1080 km/h in the eastward direction at a constant altitude in the northern hemisphere where the vertical component of earth's magnetic field is 1.75×10^-5 T. Find the emf that develops between the tips of the wings.
Solution:
Given,
length of wingspan, l = 40 m
speed, v = 1080 km/h = 300 m/s
magnetic field, B = 1.75×10^-5 T
emf developed between the tips of the wings, ε = ?

The emf developed between the tips of the wings is, \[\begin{align*} \epsilon& = Blv\\ &=1.75 \times 10^{-5} \times 40 \times 300\\ \therefore \epsilon & = 0.21 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the emf developed between the tips of the wings is 0.21 V .


A long solenoid of 1000 turns and cross sectional area 2×10^-3 m² carries a current of 2 A and produces a flux density 52×10^-3 T inside it. Calculate the self inductance of the coil.
Solution:
Given,
number of turns, N = 1000
cross sectional area, A = 2×10^-3 m²
current, I = 2 A
flux density, B = 52×10^-3 T
self inductance of the coil, L = ?

Now, the magnetic flux linked with the coil can be expressed in terms of self inductance of the coil as, \[\begin{align*} \phi&=LI\\ L&=\frac{\phi}{I}\\ &=\frac{NBA}{I} \because \phi = NBA Cos\theta = NBA Cos0^\circ = NBA\\ &=\frac{1000 \times 52 \times 10^{-3} \times 2 \times 10^{-3}}{2}\\ \therefore L&= 0.052 \hspace{0.1cm} \text{H}\\ \end{align*}\]
Thus, the self inductance of the coil is 0.052 H.

A plane circular coil has 200 turns and its radius is 0.10 m. It is connected to a battery. After switching on the circuit a current of 2 A is set up in the coil. Calculate the energy stored in the coil.
Solution:
Given,
number of turns, N = 200
radius of circular coil, r = 0.10 m
current, I = 2 A
energy stored in the coil , E = ?

The energy stored in the coil is given by, \[E=\frac{1}{2} LI^2\] At first, we need to calculate the self inductance of the coil, \[\begin{align*} L&=\frac{\phi}{I}\\ &=\frac{NBA}{I}\\ &=\frac{NA \times \mu_0 N I}{I \times 2r}\\ &=\frac{\mu_0 N^2 A}{2r}\\ &=\frac{\mu_0 N^2 \times \pi r^2}{2r}\\ &=\frac{4\pi \times 10^{-7} \times 200^2 \times \pi \times 0.10}{2}\\ &=7.89\times 10^{-3} \hspace{0.1cm} \text{H}\\ \end{align*}\]
So, \[\begin{align*} E&=\frac{1}{2} \times 7.89 \times 10^{-3} \times (2)^2\\ \therefore E&=0.016 \hspace{0.1cm} \text{J}\\ \end{align*}\]
Thus, the energy stored in the coil is 0.016 J .

Find the emf induced in a straight conductor of length 25 cm, on the armature of a dynamo and 12 cm from the axis, when the conductor is moving in a uniform radial magnetic field of 0.5 T. The armature is rotating at 1000 revolutions per minute.
Solution:
Given,
length of conductor, l = 25 cm = 0.25 m
distance from the axis of rotation, r = 12 cm = 0.12 m
magnetic field, B = 0.5 T
rate of rotation, f = 1000 rev / min = \(\frac{50}{3}\) rev / s
emf induced, \(\epsilon\) = ?

Now, \[\begin{align*} \epsilon& = Blv\\ &=Blr\omega\\ &=Blr \times 2\pi f\\ &=2\pi Blrf\\ &=2\pi \times 0.5 \times 0.25 \times 0.12 \times \frac{50}{3}\\ \therefore \epsilon&=1.57 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the emf induced in the conductor is 1.57 V .

A metal aircraft with a wing span of 40 m flies with a speed of 1000 km/h in a direction due east at constant altitude in a region of the northern hemisphere where the horizontal component of the earth's magnetic field is 1.6×10^-5 T and the angle of dip is 41⁰. Find the potential difference developed between the tips of the wing.
Solution:
Given,
length of wingspan, l = 40 m
speed, v = 1000 km/h = \(\frac{1000\times 1000}{3600}\) m/s = \(\frac{2500}{9}\) m/s
horizontal component of earth's magnetic field, \(B_H\) = 1.6×10^-5 T
angle of dip, \(\delta\) = 41⁰
potential difference between the tips of the wing, \(epsilon\) = ?
Let, \(B_V\) be the vertical component of earth's magnetic field.

Now, \[\begin{align*} \epsilon& = B_V lv\\ &=B_H tan\delta l v \hspace{0.1cm}\because tan\delta = \frac{B_V}{B_H}\\ &=1.6 \times 10^{-5}\times tan41^\circ \times 40 \times \frac{2500}{9}\\ \therefore \epsilon & = 0.155 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the potential difference developed between the tips of the wing is 0.155 V.

A rectangular coil of 100 turns has dimensions 15×10 cm. It is rotated at the rate of 300 revolutions per minute in a uniform magnetic field of flux density 0.6 T. Calculate the maximum emf induced in it.
Solution:
Given,
number of turns, N = 100
rate of rotation, f = 300 rev / min = 5 rev / s
magnetic field , B = 0.6 T
maximum emf induced ,\(\epsilon\) = ?
area, A = 150 cm² = 150×10^-4 m²


The emf induced in the coil is given by, \[\epsilon =NBA\omega sin\omega t\] For the maximum emf, \(sin \omega t = 1\). So, \[\begin{align*} \epsilon&=NBA \omega\\ &=100 \times 0.6 \times 150 \times 10^{-4} \times 2\pi f\\ &=100 \times 0.6 \times 150 \times 10^{-4} \times 2\pi \times 5\\ \therefore \epsilon & = 28.27 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the maximum emf induced in the coil is 28.27 V .

A long solenoid with 15 turns per cm has a small loop of area 2 cm² placed inside, normal to the axis of solenoid. If the current carried by the solenoid changes steadily from 2 A to 4 A in 0.1 second, what is the induced voltage in the loop, while the current is changing?
Solution:
Given,
number of turns per unit length, n = 15 / cm = 1500 / m
area, A = 2 cm² = 2×10^-4 m²
rate of change of current, \(\frac{dI}{dt}\) = \(\frac{4-2}{0.1}\) = 20 A / s
induced voltage, \(\epsilon\) = ?

Now, \[\begin{align*} \epsilon&=\frac{d\phi}{dt}\\ &=\frac{d(BA)}{dt}\\ &=\frac{d(\mu_0 n I A)}{dt}\\ &=\mu_0 An\frac{dI}{dt}\\ &=4\pi \times 10^{-7} \times 2 \times 10^{-4} \times 1500 \times 20\\ \therefore \epsilon &= 7.5 \times 10^{-6} \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the induced voltage is 7.5×10^-6 V .

Two plane coils having number of turns 1000 and 2000, and radii 5 cm and 10 cm respectively are placed co-axially in the same plane. Calculate their mutual inductance.
Solution:
Given,
number of turns in first coil, N₁ = 1000
number of turns in second coil, N₂ = 2000
radius of first coil, r₁ = 5 cm = 0.05 m
radius of second coil, r₂ = 10 cm = 0.10 m
mutual inductance, M = ?

Let the current I is passed through the outer coil. A magnetic field \(B_2=\frac{\mu_0 N_2 I}{2r_2}\) is produced inside the outer coil. The flux through each turn of inner coil is, \[\begin{align*} \phi_1&= B_2 A_1\\ &=\frac{\mu_0 N_2 I}{2r_2} \times \pi r_1^2\\ \end{align*}\]
The flux through all turns of inner coil is, \[\begin{align*} \phi_1&=N_1 B_2 A_1\\ &=N_1 \times \frac{\mu_0 N_2 I}{2r_2} \times \pi r_1^2\\ &=\frac{\mu_0 N_1 N_2 \pi r_1^2}{2r_2} \times I\\ \end{align*}\] So, the mutual inductance is, \[\begin{align*} M&=\frac{\phi_1}{I}\\ &=\frac{\mu_0 N_1 N_2 \pi r_1^2}{2r_2}\\ &=\frac{4\pi \times 10^{-7} \times 1000 \times 2000 \times \pi \times (0.05)^2}{2 \times (0.10)}\\ \therefore L&=0.0987 \hspace{0.1cm} \text{H}\\ \end{align*}\]
Thus, the mutual inductance is 0.0987 H .

A transformer has 500 turns in the primary coil and 100 turns in the secondary coil. What is the output voltage if the input voltage is 4000 volts? If the transformer is assumed to have an efficiency of 100%, what primary current is required to draw 2000 watts from the secondary?
Solution:
Given,
number of turns in the primary coil, N_p = 500
number of turns in the secondary coil, N_s = 100
input voltage, ε_p = 4000
efficiency of transformer, η = 100%
output power, P_o = 2000 W
primary current I_p = ?

Now,
For the transformer, \[\begin{align*} \frac{\epsilon_s}{\epsilon_p}&=\frac{N_s}{N_p}\\ \frac{\epsilon_s}{4000}&=\frac{100}{500}\\ \epsilon_s&=\frac{1}{5} \times 4000 \\ \therefore \epsilon_s&=800 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Since the transformer has 100 % efficiency, the input power and output power is same. \[\begin{align*} P_{i}&=P_{o}\\ \epsilon_p I_p&=P_o\\ 4000 \times I_p&=P_o\\ I_p&=\frac{2000}{4000}\\ \therefore I_p&=0.5 \hspace{0.1cm} \text{A}\\ \end{align*}\]
Thus, the primary current is 0.5 A .

When a wheel with metal spokes 1.2 m long is rotated in a magnetic field of flux density 5×10^-5 T normal to the plane of wheel, an emf of 10^-2 V is induced between the rim and axle. Find the rate of rotation of the wheel.
Solution:
Given,
length of metal spokes, r = 1.2 m
magnetic field, B = 5×10^-5 T
emf induced, ε = 10^-2 V
rate of rotation , f = ?

It is the case of motional emf. The emf induced is thus, \[\epsilon=Blv\] . Here, the velocity is an average velocity (axle is at rest and rim is moving with maximum velocity).So, \[\begin{align*} v&=\frac{0+\omega r}{2}\\ &=\frac{\omega r}{2}\\ \end{align*}\]
Now, \[\begin{align*} \epsilon&=Blv\\ &=Br \frac{\omega r}{2}\\ &=\frac{B\omega r^2}{2}\\ &=\frac{B\times 2\pi f r^2}{2}\\ \epsilon&=\pi B r^2 f\\ 10^{-2}&=\pi \times 5 \times 10^{-5} \times (1.2)^2 \times f\\ f&=\frac{10^{-2}}{\pi \times 5 \times 10^{-5} \times (1.2)^2}\\ \therefore f&=44.2 \hspace{0.1cm} \text{rev/s}\\ \end{align*}\]
Thus, the rate of rotation of the wheel is 44.2 rev/s .

A step down transformer transforms a supply line voltage 220 volts into 100 volts. Primary coil has 500 turns. The efficiency and power transmitted by the transformer are 80% and 80 kw. Find (a) the number of turns in the secondary coil (b) power supplied.
Solution:
Given,
input voltage, ε _p = 220 V
output voltage, ε _s = 100 V
number of turns in primary coil, N_p = 500
number of turns in secondary coil, N_s = ?
efficiency of transformer, η = ?
output power , P_o = 80 kw = 80000 w
input power, P_i = ?

For transformer,
\[\begin{align*} \frac{\epsilon_s}{\epsilon_p}&=\frac{N_s}{N_p}\\ \frac{100}{220}&=\frac{N_s}{500}\\ N_s&=\frac{100}{200}\times 500\\ \therefore N_s& = 250 \\ \end{align*}\] Again, \[\begin{align*} \eta&=\frac{P_o}{P_i} \times 100\%\\ 80&=\frac{80000}{P_i}\times 100\\ P_i&=\frac{80000 \times 100}{80}\\ P_i&= 100000\\ \therefore P_i&=10^5 \hspace{0.1cm} \text{w}\\ \end{align*}\]
Thus, the number of turns in the secondary coil is 250 and the power supplied by the transformer is 10⁵ W.

Updated on 24th April, 2023