Electromagnetic Induction - Class 12 Physics - Chapter 7 - Notes, NCERT Solutions & Extra Questions
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Notes - Electromagnetic Induction | Class 12 NCERT | Physics
Comprehensive Class 12 Notes: Understanding Electromagnetic Induction
Understanding the principles of electromagnetic induction is crucial for Class 12 students, particularly in the context of modern technology and practical applications. Here, you'll find comprehensive notes aimed to clarify this essential topic.
Introduction to Electromagnetic Induction
Electromagnetic induction refers to the phenomenon where an electric current is generated in a conductor due to a changing magnetic field. This concept forms the basis for many modern technologies, including electric generators and transformers.
Discovery and Historical Context
Electricity and magnetism were long viewed as unrelated phenomena. However, pivotal experiments in the early 19th century by Hans Christian Oersted, André-Marie Ampère, Michael Faraday, and Joseph Henry demonstrated their interconnected nature.
Importance in Modern Technology
Imagine a world without electricity — no lights, trains, telephones, or computers. The discovery of electromagnetic induction is the backbone of electrical engineering and modern civilisation.
Key Experiments and Discoveries
Oersted and Ampère’s Early Work
Oersted's discovery showed that electric currents create magnetic fields, and Ampère's work provided mathematical descriptions of these relationships.
Faraday’s Experiments
Michael Faraday's experiments around 1830 were groundbreaking. Faraday discovered that moving a magnet through a coil generates an electric current.
Henry’s Contributions
Joseph Henry, working independently in the USA, also made significant contributions, demonstrating electromagnetic induction through various experiments with coils.
Core Concepts of Electromagnetic Induction
Magnetic Flux
Magnetic flux (Φ) through a surface of area (A) in a magnetic field (B) is given by: [ \Phi = B \cdot A \cos \theta ] where (\theta) is the angle between (B) and (A).
Faraday’s Law of Induction
Faraday's Law states that the induced emf in a coil is equal to the rate of change of magnetic flux through it: [ \varepsilon = -N \frac{d\Phi}{dt} ]
Lenz’s Law
Lenz's Law articulates the direction of the induced current, stating that it will flow in a way to oppose the change in magnetic flux that produced it, preserving energy conservation.
graph TD;
A[Magnetic Field Change] --> B[Induced EMF] --> C[Current Opposes Change]
Calculations and Examples
The Concept of Motional EMF
When a conductor moves in a magnetic field, an emf is induced across it due to the Lorentz force acting on electrons: [ \varepsilon = B l v ]
Calculating EMF in a Moving Conductor
If a rod of length (l) moves with velocity (v) perpendicular to a magnetic field (B), the induced emf is: [ \varepsilon = B l v ]
Example Problems
Example problems can be found at the end of this article to test your understanding and reinforce the concepts.
Applications of Electromagnetic Induction
Generators and Electrical Power Generation
Generators convert mechanical energy into electrical energy using electromagnetic induction. The coil in a generator rotates within a magnetic field, altering the magnetic flux and generating an emf.
Transformers and Their Function
Transformers change the voltage of alternating current (AC) using coils and electromagnetic induction. They are essential for electrical power distribution.
graph TD;
X[High Voltage AC] --> Y[Step-Down Transformer] --> Z[Low Voltage AC for Home Use]
Induction Cooktops and Other Household Applications
Induction cooktops use electromagnetic induction to heat cookware efficiently, demonstrating the practical household applications of this phenomenon.
Advanced Concepts
Mutual Inductance
Mutual inductance ((M)) occurs when one coil induces an emf in a neighbouring coil through a changing magnetic field: [ \varepsilon_1 = -M \frac{dI_2}{dt} ]
Self-Inductance
Self-inductance ((L)) is where a coil induces an emf within itself due to a changing current: [ \varepsilon = -L \frac{dI}{dt} ] Self-inductance in a solenoid depends on its geometry and can be enhanced using materials with high magnetic permeability.
Summary and Key Equations
Recap of Key Concepts
Magnetic Flux $(\Phi)$: ( \Phi = B \cdot A \cos \theta )
Faraday’s Law: ( \varepsilon = -N \frac{d\Phi}{dt} )
Lenz’s Law: Induced current opposes the change in magnetic flux
Motional EMF: ( \varepsilon = B l v )
Mutual Inductance: ( \varepsilon_1 = -M \frac{dI_2}{dt} )
Self-Inductance: ( \varepsilon = -L \frac{dI}{dt} )
Important Equations for Quick Reference
Magnetic Flux: ( \Phi = B \cdot A \cos \theta )
Faraday’s Law: ( \varepsilon = -N \frac{d\Phi}{dt} )
Motional EMF: ( \varepsilon = B l v )
Self-Inductance: ( L = \mu_0 n^2 A l )
Tips for Students: How to Master Electromagnetic Induction
Understand the Core Principles: Grasp the basic concepts of magnetic flux, Faraday’s Law, and Lenz’s Law.
Practice Calculations: Work on problems related to motional emf and inductance.
Explore Practical Applications: Study how generators and transformers utilize electromagnetic induction.
Frequently Asked Questions
Definition and Basics
Q: What is electromagnetic induction?
A: The process of generating an electric current by changing the magnetic field around a conductor.
Q: Who are the key scientists associated with its discovery?
A: Michael Faraday and Joseph Henry.
Laws and Applications
Q: What does Faraday’s Law state?
A: It states that the induced emf is directly proportional to the rate of change of magnetic flux.
Q: How do generators use electromagnetic induction?
A: Generators convert mechanical energy into electrical energy by rotating coils within a magnetic field.
Calculation Techniques
Q: How is motional emf calculated?
A: $ \varepsilon = B l v $, where (B) is the magnetic field, (l) is the length of the conductor, and (v) is the velocity.
This article aims to provide Class 12 students with a comprehensive understanding of electromagnetic induction, preparing them for both exams and practical applications in technology.
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Notes - Flashcards - Electromagnetic Induction | Class 12 NCERT | Physics
NCERT Solutions - Electromagnetic Induction | NCERT | Physics | Class 12
Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f).
Let's analyze each figure in Fig. 6.15 with the application of Lenz's Law:
(a)
- The magnet is being moved toward the coil.
- According to Lenz’s Law, the coil will generate a current that opposes the increase in magnetic flux.
- This means the induced current in the coil will create a magnetic field opposing the approaching North pole.
- This results in a counterclockwise induced current when observed from the approaching North pole.
(b)
- The magnet is being pushed away from the coil.
- According to Lenz's Law, the coil will generate a current that opposes the decrease in magnetic flux.
- This means that the induced current in the coil will attempt to keep the magnetic field constant by creating a North pole facing the retreating North pole of the magnet.
- This results in a clockwise induced current when observed from the retreating North pole.
(c)
- The key of coil X is just closed, so the current starts flowing through coil X.
- The sudden increase in current through coil X will generate a changing magnetic field, inducing a current in coil Y.
- According to Lenz’s Law, coil Y will generate a current that opposes the increase in magnetic flux due to coil X.
- This will induce a counterclockwise current in coil Y as seen from coil X’s point of view.
(d)
- The rheostat setting is being changed, causing a change in the current and thus magnetic flux in coil X.
- This change will induce a current in coil Y to oppose the change in magnetic flux.
- If the current in coil X is increasing, the induced current in coil Y will be counterclockwise when viewed from coil X’s position.
- If the current is decreasing, the induced current in coil Y will be clockwise when viewed from coil X’s position.
(e)
- The key of coil X is just released, causing the current through coil X (and its magnetic field) to decrease rapidly.
- Coils X and Y are closely coupled, so a changing magnetic field in X will induce a current in Y.
- According to Lenz’s Law, coil Y will generate a current that opposes the decrease in magnetic flux, creating a clockwise induced current when viewed from coil X’s position.
(f)
- The current in the loop is decreasing at a steady rate.
- According to Lenz's Law, the resultant decrease in the magnetic field will induce a current that opposes this decrease.
- This current will circulate in a direction that attempts to maintain the original magnetic flux.
- Therefore, the induced current will be clockwise if viewed such that the existing magnetic field points upwards through the loop.
Summary Table:
Figure | Induced Current |
---|---|
(a) | Counterclockwise |
(b) | Clockwise |
(c) | Counterclockwise |
(d) | Counterclockwise (if increasing) Clockwise (if decreasing) |
(e) | Clockwise |
(f) | Clockwise |
Use Lenz's law to determine the direction of induced current in the situations described by Fig. 6.16:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
(a) A wire of irregular shape turning into a circular shape
As the wire changes shape from irregular to circular, the area enclosed by the wire increases, leading to an increase in the magnetic flux through the loop (given that the magnetic field is perpendicular and directed into the page).
According to Lenz's law, the induced current will flow in a direction that opposes this change in flux. Since the flux is increasing, the induced current will produce a magnetic field directed out of the page to counter the increasing flux.
To produce such a magnetic field, the induced current must flow in a counterclockwise direction when viewed from above.
(b) A circular loop being deformed into a narrow straight wire
As the circular loop is deformed into a straight wire, the area enclosed by the loop decreases, leading to a reduction in the magnetic flux through the loop (given that the magnetic field is perpendicular and directed into the page).
According to Lenz's law, the induced current will flow in a direction that opposes this change in flux. Since the flux is decreasing, the induced current will produce a magnetic field directed into the page to oppose the reduction of flux.
To create such a magnetic field, the induced current must flow in a clockwise direction when viewed from above.
Summary
- (a) Counterclockwise current to oppose the increase in flux.
- (b) Clockwise current to oppose the decrease in flux.
These directions of the induced currents satisfy Lenz's Law by opposing the respective changes in the magnetic flux through the loops.
A long solenoid with 15 turns per $\mathrm{cm}$ has a small loop of area $2.0 \mathrm{~cm}^{2}$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0 \mathrm{~A}$ to $4.0 \mathrm{~A}$ in $0.1 \mathrm{~s}$, what is the induced emf in the loop while the current is changing?
The induced emf in the loop while the current is changing is:
$$ \varepsilon = 7.54 \times 10^{-6} , \text{V} $$
or equivalently,
$$ \varepsilon = 7.54 , \mu \text{V} $$
So, the induced emf in the loop is approximately $7.54 \mu\text{V}$.
A rectangular wire loop of sides $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \mathrm{~T}$ directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is $1 \mathrm{~cm} \mathrm{~s}^{-1}$ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Given Data
- Dimensions of the rectangular loop: (8 , \text{cm} \times 2 , \text{cm})
- Magnetic field, (B): (0.3 , \text{T})
- Velocity of the loop, (v): (1 , \text{cm/s} = 0.01 , \text{m/s})
Formula
The induced emf ((\varepsilon)) when a loop moves perpendicularly out of a uniform magnetic field is given by: $$ \varepsilon = B \cdot l \cdot v $$ where (l) is the length of the side perpendicular to the direction of motion.
(a) When the loop moves normal to the longer side
In this case, (l = 8 , \text{cm} = 0.08 , \text{m})
Using the formula: $$ \varepsilon = 0.3 , \text{T} \times 0.08 , \text{m} \times 0.01 , \text{m/s} $$ $$ \varepsilon = 2.4 \times 10^{-4} , \text{V} $$
Duration of Induced Voltage
To find the duration, we need to determine how long it takes for the entire loop to exit the magnetic field region. Since the loop is moving normal to the longer side: $$ \text{Time} = \frac{\text{Length of the shorter side}}{\text{Velocity}} = \frac{2 , \text{cm}}{1 , \text{cm/s}} = 2 , \text{s} $$
(b) When the loop moves normal to the shorter side
In this case, (l = 2 , \text{cm} = 0.02 , \text{m})
Using the formula: $$ \varepsilon = 0.3 , \text{T} \times 0.02 , \text{m} \times 0.01 , \text{m/s} $$ $$ \varepsilon = 6 \times 10^{-5} , \text{V} $$
Duration of Induced Voltage
To find the duration, we need to determine how long it takes for the entire loop to exit the magnetic field region. Since the loop is moving normal to the shorter side: $$ \text{Time} = \frac{\text{Length of the longer side}}{\text{Velocity}} = \frac{8 , \text{cm}}{1 , \text{cm/s}} = 8 , \text{s} $$
Summary
-
Normal to the longer side:
- Induced emf, (\varepsilon): (2.4 \times 10^{-4} , \text{V})
- Duration: (2 , \text{s})
-
Normal to the shorter side:
- Induced emf, (\varepsilon): (6 \times 10^{-5} , \text{V})
- Duration: (8 , \text{s})
A $1.0 \mathrm{~m}$ long metallic rod is rotated with an angular frequency of $400 \mathrm{rad} \mathrm{s}^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of $0.5 \mathrm{~T}$ parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
The emf developed between the center and the ring is 100 V.
So, the final induced emf is:
$$ \varepsilon = 100 \text{ V} $$
A horizontal straight wire $10 \mathrm{~m}$ long extending from east to west is falling with a speed of $5.0 \mathrm{~m} \mathrm{~s}^{-1}$, at right angles to the horizontal component of the earth's magnetic field, $0.30 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
The instantaneous value of the emf induced in the wire is: $$\varepsilon = 1.5 , \text{mV}$$
(b) Direction of the EMF
Lenz's law states that the direction of the emf induced in a conductor moving through a magnetic field is such that it opposes the change in magnetic flux.
In this case, since the wire is falling vertically downwards and the magnetic field is horizontal from east to west, we can use the right-hand rule to determine the direction of the emf. Point your thumb in the direction of the wire's velocity (downwards) and your fingers in the direction of the magnetic field (east to west), the palm will face the direction of the induced current.
Therefore, the direction of the emf will be from south to north along the wire.
(c) Which End of the Wire is at Higher Electrical Potential?
Given that the emf direction is from south to north, the north end of the wire will be at a higher electrical potential compared to the south end, following the conventional current direction.
So, the west end of the wire will be at a higher electrical potential.
Current in a circuit falls from $5.0 \mathrm{~A}$ to $0.0 \mathrm{~A}$ in $0.1 \mathrm{~s}$. If an average emf of $200 \mathrm{~V}$ induced, give an estimate of the self-inductance of the circuit.
Given the values and calculations, the estimated self-inductance ( L ) of the circuit is:
$$ L = -4 \ \text{H} $$
Since inductance cannot be negative, we consider the magnitude:
$$ L = 4 \ \text{H} $$
Thus, the self-inductance of the circuit is 4 henry.
6.8 A pair of adjacent coils has a mutual inductance of $1.5 \mathrm{H}$. If the current in one coil changes from 0 to $20 \mathrm{~A}$ in $0.5 \mathrm{~s}$, what is the change of flux linkage with the other coil?
The change in flux linkage with the other coil is:
$$ \Delta \Phi = M \Delta I = 1.5 , \mathrm{H} \times 20 , \mathrm{A} = 30 , \mathrm{Wb} $$
So, the change in flux linkage with the other coil is 30 Weber (Wb).
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On a cylindrical rod, two coils are wound one over the other. If the inductance of each coil is 0.1 H, the mutual inductance of the pair of coils is:
A) 0.01 H
B) 0.1 H
C) 1 mH
D) 1 µH
The correct option is B.
Given:
Inductance of each coil ($ L_1 $ and $ L_2 $) is $0.1 , \text{H}$.
Assuming:
Perfect coupling between the coils, thus the coupling coefficient ($ K $) is $1$.
The formula for mutual inductance ($ M $) when coils are perfectly coupled is: $$ M = K \sqrt{L_1 L_2} $$
Substituting the given values: $$ M = 1 \cdot \sqrt{0.1 \times 0.1} = \sqrt{0.01} = 0.1 , \text{H} $$
Therefore, the mutual inductance of the pair of coils is 0.1 H, which corresponds to option B.
Electromagnetic induction refers to the:
A. charging of a body with a positive charge
B. production of current by relative motion between a magnet and a coil
C. rotation of the coil of an electric motor
D. generation of magnetic field due to a current-carrying solenoid
The correct choice is B: production of current by relative motion between a magnet and a coil.
Electromagnetic induction is a phenomenon in which an electrical current is generated by changing the magnetic field around a conductor. This can occur either when a conductor moves through a stationary magnetic field or when the magnetic field around a stationary conductor changes. This interaction induces an electromotive force (emf) across the conductor, ultimately resulting in current flow, if the circuit is closed. Thus, this process aligns directly with option B.
An electric bulb is designed to consume $55 \mathrm{~W}$ when operated at 110 volts. It is connected to a $220 \mathrm{~V}, 50 \mathrm{~Hz}$ line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage?
To find the inductance $L$ of the choke coil for operating a 55 W, 110 V bulb correctly when connected to a 220 V, 50 Hz line, we need to follow these steps:
-
Calculate the resistance of the bulb: The resistance $R$ can be calculated using the power formula: $$ P = \frac{V^2}{R} $$ Rearranging for R and substituting the given values: $$ R = \frac{V^2}{P} = \frac{110^2}{55} = 220 , \Omega $$
-
Determine the angular frequency $\omega$: Given the frequency $f = 50 , \text{Hz}$, the angular frequency $\omega$ is given by: $$ \omega = 2\pi f = 2\pi \times 50 = 100\pi , \text{rad/s} $$
-
Model the circuit: The bulb and choke coil are in series, with a total impedance that includes both the resistor (bulb) and the inductor (choke coil). The impedance $Z$ is: $$ Z = \sqrt{R^2 + (\omega L)^2} $$
-
Calculate the current in the circuit: With the total voltage $V = 220 , \text{V}$ across the series combination, the current $I$ through the circuit is: $$ I = \frac{V}{Z} = \frac{220}{\sqrt{R^2 + (\omega L)^2}} $$
-
Required voltage across the bulb/resistor ($R$): Since the bulb should receive the correct voltage of 110 V: $$ V_R = IR = \frac{220 \times 220}{\sqrt{220^2 + (100\pi L)^2}} = 110 $$ Simplify and solve for $L$: $$ 220 = \sqrt{220^2 + (100\pi L)^2} $$ $$ (220)^2 + (100\pi L)^2 = (440)^2 $$ $$ 48400 + 10000 \pi^2 L^2 = 193600 $$ $$ 10000 \pi^2 L^2 = 145200 \implies L^2 = \frac{145200}{10000 \pi^2} $$ $$ L = \sqrt{\frac{145200}{10000 \pi^2}} \approx 1.2135 , \text{H} $$ We round this to: $$ L \approx 1.2 , \text{H} $$ This inductance ( L = 1.2 , \text{H} ) is what the choke coil should possess to reduce the 220 V supply to a suitable 110 V across the 55 W bulb, ensuring it operates correctly.
Name the coil of which the wire is thicker in a (i) step-up, (ii) step-down transformer. Give a reason for your answer.
Solution
-
Step-Up Transformer
- A transformer that increases the voltage is called a step-up transformer (i.e., $ K > 1 $ or $ N_s > N_p $, where $ N_s $ and $ N_p $ are the number of turns in the secondary and primary coils, respectively).
- In this setup, the current in the secondary coil is less than that in the primary coil. To support the higher primary current, the primary coil is made of thicker wire, which is capable of handling this higher current demand.
-
Step-Down Transformer
- A step-down transformer reduces the voltage while maintaining constant power output.
- This decrease in voltage leads to an increase in current ($ I $) on the secondary side (based on the formula $ P = V \times I $, where $ P $ is power, $ V $ is voltage, and $ I $ is current). To manage this increased current, the secondary coil uses thicker wire.
- Additionally, using a thicker wire helps to reduce $ I^2R $ losses, as thicker wires generally have lower resistance.
-
Conclusion
- In summary, in a step-up transformer, the primary coil uses thicker wires, whereas in a step-down transformer, it is the secondary coil that generally requires thicker wires. Both adjustments cater to the need of efficiently handling current differences and minimizing resistive losses in their respective circuits.
A magnet is brought towards a coil rapidly in the first case and gently in the second case. The induced e.m.f. will be:
A) Lesser in the first case.
B) Greater in the first case.
C) Equal in both cases.
D) Insufficient data given.
The correct answer is B) Greater in the first case.
The phenomenon behind this is based on Faraday's Law of Electromagnetic Induction. According to Faraday's Law, an electromotive force (emf) is induced in a coil when there is a change in the magnetic flux linked with it. Mathematically, this relationship is expressed as:
$$ \text{emf} = -\frac{\Delta \Phi}{\Delta t} $$
where $ \Delta \Phi $ is the change in magnetic flux and $ \Delta t $ is the time over which the change occurs. Essentially, the strength of the induced emf is directly proportional to the rate of change of magnetic flux through the coil.
When a magnet is brought towards a coil rapidly (as in the first case), the magnetic flux through the coil changes more quickly compared to when the magnet is moved gently (as in the second case). Thus, there's a higher rate of change of magnetic flux in the first case. Consequently, the induced emf is greater when the magnet is moved rapidly than when moved gently.
"The principle of electromagnetic induction was given by:
A) Oersted
B) Faraday
C) Ampere"
The correct response is B) Faraday.
Michael Faraday introduced the principle of electromagnetic induction. According to Faraday's Law of electromagnetic induction, an electromotive force (emf) is induced in a closed circuit due to variations in the magnetic field enclosing the circuit.
A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time $t=0$, so that a time-dependent current $\mathrm{I}_{1}(t)$ starts flowing through the coil. If $\mathrm{I}_{2}(t)$ is the current induced in the ring, and $B(t)$ is the magnetic field at the axis of the coil due to them as a function of time $(t>0)$, the product $I_{2}(t) B(t)$:
Increases with time
Decreases with time
Does not vary with time
Passes through a maximum
The correct option is D: Passes through a maximum.
We employ constants such as $k_1$, $k_2$, etc., to simplify the expressions. Let's break it down step-by-step:
Current through the coil as a function of time: $$ I_{1}(t) = k_{1}\left[1-e^{-t / \tau}\right] $$
Magnetic field at the axis of the coil due to the current: $$ B(t) = k_{2} I_{1}(t) $$ Substituting $I_{1}(t)$, we get: $$ B(t) = k_{2} k_{1}\left[1-e^{-t / \tau}\right] $$
Induced current in the ring: $$ I_{2}(t) = k_{3} \frac{d B(t)}{d t} $$ Given that $B(t) = k_{2} k_{1}\left[1-e^{-t / \tau}\right]$, we need to find its derivative with respect to time $t$: $$ \frac{d B(t)}{d t} = k_{2} k_{1} \left( \frac{d}{dt} [1 - e^{-t/\tau}] \right) = k_{2} k_{1} \left(0 - \left[-\frac{1}{\tau}e^{-t/\tau}\right]\right) = k_{2} k_{1} \left(\frac{1}{\tau}e^{-t/\tau}\right) $$ Thus: $$ I_{2}(t) = k_{3} \left(k_{2} k_{1} \frac{1}{\tau}e^{-t/\tau}\right) = k_{4} e^{-t / \tau} $$
Product of the induced current and the magnetic field: $$ I_{2}(t) B(t) = k_{4} e^{-t / \tau} \cdot k_{2} k_{1}\left[1-e^{-t / \tau}\right] = k_{5}\left[1-e^{-t / \tau}\right] e^{-t / \tau} $$
Observing this product:
At $t = 0$:$$ I_{2}(0) B(0) = k_{5} \left(1 - e^{0 / \tau}\right) e^{0 / \tau} = k_{5} \cdot 0 \cdot 1 = 0 $$
As $t \to \infty$:$$ I_{2}(\infty) B(\infty) = k_{5} \left(1 - e^{-\infty / \tau}\right) e^{-\infty / \tau} = k_{5} \cdot 1 \cdot 0 = 0 $$
Hence, $I_{2}(t) B(t)$ is zero at both $t=0$ and as $t \to \infty$. This product is positive for other values of $t$, indicating that it must pass through a maximum at some intermediate value of $t$.
A rectangular coil of 20 turns and cross-sectional area 25 cm^2 has a resistance of 100 ohms. If a magnetic field which is perpendicular to the plane of the coil changes at a rate of 1000 T/s, the current in the coil is:
A) 1 A
B) 50 A
C) 0.5 A
D) 5 A
The correct option is C: $$0.5 , \text{A}$$
Explanation:
Faraday's Law of Electromagnetic Induction states that the emf (electromotive force) generated in a coil is given by: $$ E = n \frac{d(BA)}{dt} $$ where:
( E ) is the induced emf
( n ) is the number of turns in the coil
( B ) is the magnetic field
( A ) is the area of the coil
Given:
Number of turns, ( n = 20 )
Cross-sectional area, ( A = 25 , \text{cm}^2 = 25 \times 10^{-4} , \text{m}^2 )
Rate of change of magnetic field, ( \frac{dB}{dt} = 1000 , \text{T/s} )
The induced emf is: $$ E = 20 \times 1000 \times 25 \times 10^{-4} , \text{V} $$ Simplifying this, we get: $$ E = 50 , \text{V} $$
Ohm's Law states that the current ( I ) is given by: $$ I = \frac{E}{R} $$ where:
( E ) is the emf
( R ) is the resistance of the coil
Given:
Resistance, ( R = 100 , \Omega )
Finally, the current in the coil is: $$ I = \frac{50}{100} = 0.5 , \text{A} $$
Thus, the current in the coil is 0.5 A.
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
A 0.138 H B 1.389 H C 138.88 H D 13.89 H
The correct answer is Option D: 13.89 H.
To determine the inductance, we start with the formula for the energy stored in an inductor:
$$ U = \frac{1}{2} L I^2 $$
Given:
Energy stored, $$ U = 25 \text{ mJ} = 25 \times 10^{-3} \text{ J} $$
Current, $$ I = 60 \text{ mA} = 60 \times 10^{-3} \text{ A} $$
Substituting the values into the energy equation:
$$ 25 \times 10^{-3} = \frac{1}{2} \times L \times (60 \times 10^{-3})^2 $$
To isolate $L$, we rearrange the equation:
$$ L = \frac{2 \times 25 \times 10^{-3}}{(60 \times 10^{-3})^2} $$
Next, we perform the calculations:
$$ L = \frac{50 \times 10^{-3}}{(60 \times 10^{-3})^2} $$
$$ L = \frac{50 \times 10^{-3}}{3600 \times 10^{-6}} $$
$$ L = \frac{50 \times 10^3}{3600} $$
$$ L = \frac{500}{36} $$
$$ L = 13.89 \text{ H} $$
Thus, the inductance of the inductor is 13.89 H.
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