Electromagnetic induction | Class 12 NEB Physics | Numerical problem solution
Electromagnetic Induction Problems
Physics · EMI, Inductance & Transformers
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.
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,
\[\begin{align*} \epsilon&=-(8 \times 2 + 5)\\ \therefore \epsilon& = -21\hspace{0.1cm}\text{V} \end{align*}\]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. Calculate the emf induced across its length.
The emf induced in a straight current carrying conductor is,
\[\begin{align*} \epsilon&=Blv\sin\theta\\ &=10 \times 0.25 \times 10 \times \sin 45^\circ\\ &=17.67 \hspace{0.1cm} \text{V} \end{align*}\]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?
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,
\[\begin{align*} \Delta \phi&=NBA\cos 0^\circ - NBA\cos 180^\circ\\ &=NBA+NBA\\ &=2NBA\\ &=2 \times 100 \times 3\times 10^{-3} \times 2 \times 10^{-3}\\ \Delta \phi&=1.2\times 10^{-3} \hspace{0.1cm} \text{Wb} \end{align*}\]Now, the charge circulated through the coil is,
\[\begin{align*} \Delta q&=\frac{\Delta \phi}{R}=\frac{1.2 \times 10^{-3}}{12}\\ \therefore \Delta q&= 10^{-4} \hspace{0.1cm} \text{C} \end{align*}\]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.
The emf developed between the tips of the wings is,
\[\begin{align*} \epsilon &= Blv\\ &=1.75 \times 10^{-5} \times 40 \times 300\\ \therefore \epsilon &= 0.21 \hspace{0.1cm} \text{V} \end{align*}\]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.
The magnetic flux linked with the coil can be expressed in terms of self inductance as,
\[\begin{align*} \phi&=LI\\ L&=\frac{\phi}{I}\\ &=\frac{NBA}{I} \quad \because \phi = NBA\cos\theta = NBA\cos 0^\circ = NBA\\ &=\frac{1000 \times 52 \times 10^{-3} \times 2 \times 10^{-3}}{2}\\ \therefore L&= 0.052 \hspace{0.1cm} \text{H} \end{align*}\]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.
The energy stored in the coil is given by \(E=\frac{1}{2}LI^2\). First, calculate the self inductance:
\[\begin{align*} L&=\frac{\phi}{I}=\frac{NBA}{I}\\ &=\frac{NA \times \mu_0 N I}{I \times 2r}\\ &=\frac{\mu_0 N^2 A}{2r}=\frac{\mu_0 N^2 \times \pi r^2}{2r}\\ &=\frac{4\pi \times 10^{-7} \times 200^2 \times \pi \times 0.10}{2}\\ &=7.89\times 10^{-3} \hspace{0.1cm} \text{H} \end{align*}\]So,
\[\begin{align*} E&=\frac{1}{2} \times 7.89 \times 10^{-3} \times (2)^2\\ \therefore E&=0.016 \hspace{0.1cm} \text{J} \end{align*}\]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.
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.
Let BV be the vertical component of earth's magnetic field. Now,
\[\begin{align*} \epsilon &= B_V lv\\ &=B_H \tan\delta \cdot lv \quad \because \tan\delta = \frac{B_V}{B_H}\\ &=1.6 \times 10^{-5}\times \tan 41^\circ \times 40 \times \frac{2500}{9}\\ \therefore \epsilon &= 0.155 \hspace{0.1cm} \text{V} \end{align*}\]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.
The emf induced in the coil is given by \(\epsilon = NBA\omega\sin\omega t\). For the maximum emf, \(\sin\omega t = 1\). So,
\[\begin{align*} \epsilon&=NBA\omega\\ &=100 \times 0.6 \times 150 \times 10^{-4} \times 2\pi f\\ &=100 \times 0.6 \times 150 \times 10^{-4} \times 2\pi \times 5\\ \therefore \epsilon &= 28.27 \hspace{0.1cm} \text{V} \end{align*}\]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?
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.
Let the current I be 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,
\[\begin{align*} M&=\frac{\phi_1}{I}=\frac{\mu_0 N_1 N_2 \pi r_1^2}{2r_2}\\ &=\frac{4\pi \times 10^{-7} \times 1000 \times 2000 \times \pi \times (0.05)^2}{2 \times 0.10}\\ \therefore M&=0.0987 \hspace{0.1cm} \text{H} \end{align*}\]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?
Since the transformer has 100% efficiency, input power = output power.
\[\begin{align*} P_i&=P_o\\ \epsilon_p I_p&=P_o\\ I_p&=\frac{2000}{4000}\\ \therefore I_p&=0.5 \hspace{0.1cm} \text{A} \end{align*}\]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.
It is the case of motional emf. The emf induced is \(\epsilon=Blv\). Here, the velocity is an average velocity (axle is at rest and rim is moving with maximum velocity). So,
\[v=\frac{0+\omega r}{2}=\frac{\omega r}{2}\]Now,
\[\begin{align*} \epsilon&=Blv=Br\frac{\omega r}{2}=\frac{B\omega r^2}{2}=\frac{B\times 2\pi f r^2}{2}\\ \epsilon&=\pi B r^2 f\\ 10^{-2}&=\pi \times 5 \times 10^{-5} \times (1.2)^2 \times f\\ \therefore f&=44.2 \hspace{0.1cm} \text{rev/s} \end{align*}\]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.
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