This chapter explores the phenomenon of generating electric current from changing magnetic fields.
Key Concepts:
Electromagnetic Induction : The production of electric current in a conductor due to a change in magnetic flux.
Faraday’s Law: The magnitude of the induced electromotive force (emf) is proportional to the rate of change of magnetic flux.
that produced it.
AC Generators: Machines that convert mechanical energy into electrical energy by exploiting electromagnetic induction.
Alternating Current (AC): A current that periodically reverses its direction of flow.
Transformers: Devices that employ electromagnetic induction to modify the voltage of an alternating current.
By understanding these concepts, we can appreciate the fundamental principles behind various electrical devices and systems.
1. Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f ).
Ans :Lenz’s Law dictates the direction of the induced current in a closed loop. The provided figures illustrate the induced current direction when a bar magnet’s North pole is moved towards and away from a closed loop, respectively.
Applying Lenz’s Law, we can predict the induced current direction in the given scenarios:
(a) The induced current flows from q to r, then to p, and finally back to q.
(b) The induced current flows from p to r, then to q, and finally back to p.
(c) The induced current flows from y to z, then to x, and finally back to y.
(d) The induced current flows from z to y, then to x, and finally back to z.
(e) The induced current flows from x to r, then to y, and finally back to x.
(f): Since the magnetic field lines are parallel to the plane of the closed loop, no change in magnetic flux occurs. Therefore, no induced current is generated.
2. 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.
Ans :(a) As the wire of irregular shape is transformed into a circular shape, the area enclosed by the wire increases. This causes an increase in the magnetic flux linked with the wire. In accordance with Lenz’s law, the induced current generates a magnetic field that opposes the increase in magnetic flux. Therefore, the induced current will produce a magnetic field that opposes the original magnetic field. This can be achieved by the induced current flowing in a direction such that its magnetic field points out of the plane of the page.
(b) As the circular loop is deformed into a narrow straight wire, the area enclosed by the wire decreases. This causes a decrease in the magnetic flux linked with the wire. According to Lenz’s law, the induced current will flow in such a direction as to oppose this decrease in magnetic flux. The induced current will produce a magnetic field that reinforces the direction of the original magnetic field. This can be achieved by the induced current flowing in a direction such that its magnetic field points into the plane of the page.
3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Ans :We can use Faraday’s Law to solve this problem, which relates induced emf to the rate of change of magnetic flux.
Step 1: Determine Magnetic Field Inside the Solenoid
(B) denote magnetic field inside a solenoid given by following formula
B = μ₀nI
Where:
μ₀ = permeability of free space
(4π x 10⁻⁷ T m/A)
Number of turns packed per unit length: (n)
(1500 turns/m)
I = current flowing through the solenoid
Step 2: Determine Change in Magnetic Flux
(Φ)denote magnetic flux through the small loop is given by following formula
Φ = BA
Where:
B = magnetic field
A = area of the loop
(2.0 x 10⁻⁴ m²)
The change in magnetic flux (ΔΦ) through the loop is equal to the product of the loop’s area and the change in the magnetic field (ΔB).
ΔΦ = AΔB
Step 3: Find the Induced EMF
According to Faraday’s Law, the induced emf (ε) is the negative rate of change of magnetic flux:
ε = -dΦ/dt
Substituting the expression for ΔΦ:
ε = -A(ΔB/Δt)
We can find ΔB/Δt using the formula for the solenoid’s magnetic field and the given change in current and time:
ΔB/Δt = μ₀n(ΔI/Δt)
Substituting this into the expression for ε:
ε = -Aμ₀n(ΔI/Δt)
Now, plugging in the given values:
ε = -2.0 x 10⁻⁴ m² * 4π x 10⁻⁷ T m/A * 1500 turns/m * (2.0 A / 0.1 s)
Calculating the value:
ε ≈ -7.54 x 10⁻⁵ V
The negative sign implies that the induced electromotive force opposes the variation in magnetic flux.
Therefore, the magnitude of the induced emf in the loop is approximately 7.54 x 10⁻⁵ volts.
4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm 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?
Ans :A rectangular loop is exiting a uniform magnetic field.. We need to find the induced emf across the cut in two scenarios: when the loop moves out along its longer side and when it moves out along its shorter side.
Solution:
Faraday’s Law of Electromagnetic Induction:
The electromotive force induced in a loop is equivalent to the time rate of change of magnetic flux linked with the loop.
ε = -dΦ/dt
The time rate of change of magnetic flux for a rectangular loop moving orthogonally to a magnetic field B is:
dΦ/dt = B * dA/dt
Case (a): Loop moving along the longer side
The enclosed area within the loop is shrinking at a rate of…
dA/dt = B * v * (shorter side)
So, the induced emf is:
ε = B * v * (shorter side) = 0.3 T * 0.01 m/s * 0.02 m = 6 × 10⁻⁵ V
The induced voltage will last for the time it takes the longer side to move out of the magnetic field:
Time = Length / Velocity = 0.08 m / 0.01 m/s = 8 s
Case (b): Loop moving along the shorter side
Similarly, in this case:
dA/dt = B * v * (longer side)
So, the induced emf is:
ε = B * v * (longer side)
= 0.3 T * 0.01 m/s * 0.08 m
= 2.4 × 10⁻⁴ V
The induced voltage will last for the time it takes the shorter side to move out of the magnetic field:
Time = Length / Velocity = 0.02 m / 0.01 m/s = 2 s
Therefore:
In case (a), the induced emf is 6 × 10⁻⁵ V and lasts for 8 seconds.
In case (b), the induced emf is 2.4 × 10⁻⁴ V and lasts for 2 seconds.
5. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad 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 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Ans :As the rod rotates, it sweeps out a circular area. This change in area within the magnetic field induces an electromotive force (EMF) within the rod.
Determine induced EMF (ε) with following formula
ε = (1/2)BωL²
Where:
B = magnetic field strength
(0.5 T)
ω= angular frequency
(400 rad/s)
L = length of the rod
(1.0 m)
Substituting the given values:
ε = (1/2) * 0.5 T * 400 rad/s * (1.0 m)² = 100 V
Hence, a 100-volt electromotive force (EMF) is produced between the center and the ring.
6. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m S-1 , at right angles to the horizontal component of the earth’s magnetic field, 0.30 ´ 10-4 Wb 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?
Ans :Analyzing the Induced EMF in the Falling Wire
Understanding the Problem:
A horizontal wire traverses the Earth’s magnetic field. We need to find the induced emf, its direction, and the polarity of the ends of the wire.
Solution:
a) Instantaneous Value of the Induced EMF:
The induced emf (ε) in a conductor moving with velocity (v) perpendicular to a magnetic field (B) is given by:
ε = B * L * v
Where:
B = magnetic field
(0.30 × 10⁻⁴ T)
L = length of the conductor
(10 m)
v = velocity of the conductor
(5.0 m/s)
Substituting the values:
ε = 0.30 × 10⁻⁴ T * 10 m * 5.0 m/s = 1.5 × 10⁻³ V
b) Direction of the EMF:
To find the direction of the induced emf, we can use Fleming’s Right-Hand Rule. Point your thumb downwards (direction of the conductor’s motion), your index finger towards the north (direction of the magnetic field), and your middle finger will indicate the direction of the induced current.
The induced current flows west to east, making the west end of the wire positively charged.
c) Polarity of the Ends of the Wire:
As determined above, the west end of the wire will be at a higher potential. Therefore, the west end is positively charged, and the east end is negatively charged.
In summary:
The induced emf is 1.5 × 10⁻³ V.
The direction of the emf is from west to east.
The west terminus of the wire exhibits a higher electrical potential.
7. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Ans :We can utilize the formula for the induced electromotive force (EMF) in a coil due to self-inductance:
ε = -L(dI/dt)
Where:
ε= induced EMF
(200 V)
L = self-inductance in Henrys
dI/dt = rate of change of current
The rate of change of current (dI/dt) can be calculated as:
dI
———
dt
= (Final current – Initial current)
———————————————–
time interval
Substituting the given values:
dI
———-
dt
(0.0 A – 5.0 A)
= ———————————
0.1 s = -50 A/s
By substituting the values of ε and dI/dt into the self-inductance formula and solving for L, we get:
200 V = -L * (-50 A/s)
L =
200 V
———–
50 A/s
= 4 H
Hence, the circuit’s self-inductance is 4 Henries.
8. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Ans :We know that the mutual inductance (M) between two coils is defined as the ratio of the change in magnetic flux (ΔΦ) linked with one coil to the change in current (ΔI) flowing through the other coil.
Mathematically,
M = ΔΦ / ΔI
Given:
M
= 1.5 H
ΔI
= 20 A – 0 A
= 20 A
Δt
= 0.5 s
We need to find ΔΦ.
Rearranging the formula:
ΔΦ = M * ΔI
Substituting the values:
ΔΦ = 1.5 H * 20 A = 30 Weber
Therefore, the change in flux linkage with the other coil is 30 Weber.
