Class 12 - Unit # 19 : Electromagnetic Induction -Solved Numericals
Physics Problems - Electromagnetism & Induction
Problem 1: Mutual Inductance and Rate of Change of Current
Two coils are placed adjacent to each other, and a change in current in the first coil induces an emf of 0.5 V in the second coil. If the mutual inductance is 0.2 H, calculate the rate of change of current in the first coil.
Data:
Induced Electromotive Force (εs) = 0.5 V
Mutual Inductance (M) = 0.2 H
Rate of Change of Current (ΔIp / Δt) = ?
Solution:
According to the definition of mutual inductance, the induced electromotive force in the secondary coil is proportional to the rate of change of current in the primary coil:
εs = M · (ΔIp / Δt)
Rearranging the formula to solve for the rate of change of current:
ΔIp / Δt = εs / M
Substituting the given values into the equation:
ΔIp / Δt = 0.5 / 0.2
ΔIp / Δt = 2.5 A/s
Result: The rate of change of current in the first coil is 2.5 A/s.
Problem 2: Self-Inductance and Induced EMF
A coil with an inductance of 0.5 H experiences a rate of change of current of 2 A/s. Calculate the induced EMF in the coil.
Data:
Self Inductance (L) = 0.5 H
Rate of Change of Current (ΔI / Δt) = 2 A/s
Induced EMF (ε) = ?
Solution:
According to Faraday's Law of Induction applied to self-inductance (Lenz's Law accounts for the negative sign indicating opposition):
ε = -L · (ΔI / Δt)
Substituting the given values into the equation:
ε = -0.5 · 2
ε = -1 V
Result: The induced EMF in the coil is -1 V.
Problem 3: Finding Mutual Inductance
Two coils are placed close to each other. If a change in current of 3 A/s in the first coil induces an EMF of 4 V in the second coil, calculate the mutual inductance.
Data:
Rate of Change of Current (ΔIp / Δt) = 3 A/s
Induced EMF (εs) = 4 V
Mutual Inductance (M) = ?
Solution:
According to the definition of mutual inductance (taking magnitude):
εs = M · (ΔIp / Δt)
Rearranging the equation to solve for the mutual inductance (M):
M = εs / (ΔIp / Δt)
Substituting the given values:
M = 4 / 3
M = 1.33 H
Result: The mutual inductance is equal to 1.33 H.
Problem 4: Step-Up Transformer Secondary Voltage
A transformer has 200 turns in the primary coil and 400 turns in the secondary coil. If the primary voltage is 120 V, calculate the secondary voltage for a step-up transformer.
Data:
Number of Turns in Primary (Np) = 200 turns
Number of Turns in Secondary (Ns) = 400 turns
Voltage in Primary (Vp) = 120 V
Voltage in Secondary (Vs) = ?
Solution:
Using the fundamental transformer ratio equation:
Vs / Vp = Ns / Np
Rearranging to solve explicitly for secondary voltage (Vs):
Vs = Vp · (Ns / Np)
Substituting the values into the relation:
Vs = 120 · (400 / 200)
Vs = 120 · 2
Vs = 240 V
Result: The secondary voltage for a step-up transformer is 240 V.
Problem 5: Energy Stored in an Inductor
An inductor with an inductance of 0.02 H has a current flowing through it of 2 A. Calculate the energy stored in the inductor.
Data:
Inductance (L) = 0.02 H
Electric Current (I) = 2 A
Energy Stored (Um) = ?
Solution:
The potential energy stored within the magnetic field of an inductor is given by the formula:
Um = ½ · L · I2
Substituting the values into the equation:
Um = ½ · (0.02) · (2)2
Um = 0.01 · 4
Um = 0.04 J
Result: The energy stored in the inductor is 0.04 J.
Problem 6: Calculating Inductance from Stored Energy
A coil stores energy in the form of magnetic potential energy of 0.2 J when it carries a current of 2 A. Calculate the inductance of the coil.
Data:
Energy Stored (Um) = 0.2 J
Electric Current (I) = 2 A
Inductance (L) = ?
Solution:
The equation for energy stored in an inductor is:
Um = ½ · L · I2
Rearranging the equation to solve for inductance (L):
L = (2 · Um) / I2
Substituting the given values into the formula:
L = (2 · 0.2) / (2)2
L = 0.4 / 4
L = 0.1 H
Result: The inductance of the coil is 0.1 H.
Problem 7: AC Generator Voltages (RMS to Peak)
An AC generator produces an alternating current with a root-mean-square (RMS) voltage of 240 V. If the frequency of the generated AC is 50 Hz, calculate the peak value of the voltage.
Data:
RMS Voltage (Vrms) = 240 V
Frequency (f) = 50 Hz
Peak Voltage (Vo) = ?
Solution:
The mathematical relationship between the root-mean-square voltage and the peak voltage value is:
Vrms = Vo / √2
Rearranging the expression to compute the peak voltage (Vo):
Vo = Vrms · √2
Substituting the known parameters (√2 ≈ 1.414):
Vo = 240 · 1.4142
Vo ≈ 339.41 V
Result: The peak value of the voltage is approximately 339 V.
Problem 8: Step-Down Transformer Secondary Voltage
A transformer has 1000 turns in its primary coil and 200 turns in its secondary coil. If the primary voltage is 120 V, calculate the secondary voltage for a step-down transformer.
Data:
Number of Turns in Primary (Np) = 1000 turns
Number of Turns in Secondary (Ns) = 200 turns
Voltage in Primary (Vp) = 120 V
Voltage in Secondary (Vs) = ?
Solution:
Applying the standard transformer expression:
Vs / Vp = Ns / Np
Rearranging to find secondary voltage (Vs):
Vs = Vp · (Ns / Np)
Substituting the given numbers:
Vs = 120 · (200 / 1000)
Vs = 120 · 0.2
Vs = 24 V
Result: The secondary voltage for a step-down transformer is 24 V.
Problem 9: Motional EMF Induced in a Conductor
A conductor of length 0.4 m moves at a velocity of 5 m/s perpendicular to a magnetic field of 0.3 T. Calculate the motional EMF induced in the conductor.
Data:
Length of Conductor (L) = 0.4 m
Velocity (v) = 5 m/s
Angle (θ) = 90° (since velocity is perpendicular to the field)
Magnetic Field Strength (B) = 0.3 T
Motional EMF (ε) = ?
Solution:
The general formula for motional electromagnetic force induced across a moving conductor path is:
ε = B · L · v · sin(θ)
Since the motion is exactly perpendicular (sin(90°) = 1):
ε = 0.3 · 0.4 · 5 · sin(90°)
ε = 0.3 · 2 · 1
ε = 0.6 V
Result: The motional EMF induced in the conductor is 0.6 V.
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