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Physics
UTME / JAMB 2024
Electromagnetic Field
Solution
If a charge ion goes through a combined electric field E and magnetic field B, the resultant emergent velocity of the ion is
A
\(\frac{E}{B}\)
B
EB
C
\(\frac{B}{E}\)
D
E-B
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The resultant emergent velocity of a charged ion moving through combined electric and magnetic fields can be derived from the condition where the electric force equals the magnetic force. This gives us the formula for the velocity v:
q E = qvB
v = \(\frac{E}{B}\) (q will cancel out)
NOTE: When both fields are present, for the ion to move without deflection, the electric force must equal the magnetic force.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70435)
Physics
UTME / JAMB 2024
Electromagnetic Field
Solution
An ideal transformer has
A
Maximum primary resistance
B
No flux leakage
C
Hysteresis
D
Eddy current
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Maximum primary resistance: An ideal transformer is assumed to have zero resistance in its windings.
No flux leakage: This is the key feature of an ideal transformer; all the magnetic flux produced by the primary winding links with the secondary winding.
Hysteresis: While hysteresis losses can occur in real transformers due to the magnetic properties of the core material, an ideal transformer assumes no losses.
Eddy current: Similar to hysteresis, eddy currents are not present in an ideal transformer, which assumes a perfect core with no energy losses
Source: WAEC / JAMB / NECO Classroom Archive (Record #70483)
Physics
UTME / JAMB 2024
Electromagnetic Field
Solution
Which of the following measuring instruments operates based on the heating effect of electric current?
A
moving iron ammeter
B
hot wire ammeter
C
galvanometer
D
electric heater
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Hot wire ammeters measure current by detecting the heat produced in a wire due to the electric current flowing through it.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70674)
Physics
UTME / JAMB 2024
Electromagnetic Field
Solution
As per Faraday's laws of electromagnetic induction, an e.m.f is induced in a conductor whenever
A
lies parallel to the magnetic flux
B
lies in a magnetic field
C
lies out side but close to a magnetic field
D
cuts magnetic flux
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Cuts magnetic flux: According to Faraday's law, when a conductor cuts through magnetic field lines (flux), an e.m.f is induced.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70692)
Physics
WAEC / WASSCE 2024
Electromagnetic Field
Solution
Which of the following modes is the most economical method of transmitting electrical power over a long distance?
A
Alternating current at high voltage and low current.
B
Alternating current at low voltage and high current.
C
Direct current at high voltage and low current.
D
Direct current at low voltage and high current.
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
When electricity is transmitted over long distances, power losses occur mainly due to the resistance of the transmission lines. These losses are proportional to the square of the current (I\(^2\)R losses). By using high voltage, the current can be kept low, significantly reducing these losses
Source: WAEC / JAMB / NECO Classroom Archive (Record #72071)
Physics
UTME / JAMB 2024
Electric Field
Solution
Find the value of a capacitor with voltage 5V and 30C.
A
7f
B
6f
C
15f
D
0.17f
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The capacitance of a capacitor is given by: C = \(\frac{Q}{V}\)
where Q = quantity of charges, and V = voltage
Q = 30c and V = 5v
C = \(\frac{Q}{V}\) = C = \(\frac{30}{5}\) = 6 farad
Source: WAEC / JAMB / NECO Classroom Archive (Record #70356)
Physics
UTME / JAMB 2024
Electric Field
Solution
The charge of magnitude 1.6 x 10 \(^{-19}\)C is placed in a uniform electric field of intensity 1200Vm\(^{-1}\). Calculate its acceleration, if the mass of the charge is 9.1 x 10\(^{-31}\)kg
A
2.11 x 10\(^{14}\)ms\(^{-2}\)
B
2.11 x 10\(^{13}\)ms\(^{-2}\)
C
1.20 x 10\(^{14}\)ms\(^{-2}\)
D
1.20 x 10\(^{13}\)ms\(^{-2}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Electric field intensity (E) = \(\frac{\text{force}}{\text{charge}}\) = \(\frac{\text{F}}{\text{q}}\)
Given: E = 1200Vm\(^{-1}\), q = 1.6 x 10 \(^{-19}\)C, m = 9.1 x 10\(^{-31}\)kg
F = E x q
F = 1.6 x 10\(^{-19}\) x 1200 = 1.92 x 10\(^{-16}\)N
F = ma
a = \(\frac{F}{m}\) = \(\frac{1.92\times 10^{-16}}{9.1\times 10^{-31}}\) \(\approxeq\) 2.11 x 10\(^{14}\)ms\(^{-2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70510)
Physics
UTME / JAMB 2024
Electric Field
Diagram
Solution
The energy stored in the above capacitor is
Attached Figure / Diagram
A
2.25J
B
6.25J
C
18.25J
D
24.25J
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The energy stored in the capacitor = \(\frac{1}{2}\)\(\frac{q^2}{C}\)
Where C = 2F, q = 3C
= \(\frac{1}{2}\)\(\frac{3^2}{2}\) = \(\frac{9}{4}\) = 2.25J
Source: WAEC / JAMB / NECO Classroom Archive (Record #70518)
Physics
UTME / JAMB 2024
Electric Field
Solution
Calculate the value of electric field intensity due to a charge of 4μC if the force due to the charge is 8N
A
2.0 x 10\(^6\)NC\(^{-1}\)
B
3.0 x 10\(^6\)NC\(^{-1}\)
C
4.0 x 10\(^6\)NC\(^{-1}\)
D
8.0 x 10\(^6\)NC\(^{-1}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Electric field intensity(E) = \(\frac{\text{Force}}{\text{charge}}\) = \(\frac{F}{q}\)
q = 4\muC = 4.0 x 10\(^{-6}\)C
F = 8N
E = \(\frac{8}{4.0 \times 10^{-6}}\) = 2.0 x 10\(^6\)NC\(^{-1}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70680)
Physics
UTME / JAMB 2024
Electric Field
Solution
Two capacitors of 0.0003μF and 0.0006μF are connected in series, find their combined capacitance.
A
0.0009μF
B
0.0002μF
C
0.0005μF
D
0.0001μF
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Combined capacitance in series = \(\frac{1}{C_T}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\)
The two capacitors have capacitance of 0.0003\muF and 0.0006\muF respectively
\(\frac{1}{C_T}\) = \(\frac{1}{0.0003}\) + \(\frac{1}{0.0006}\) = \(\frac{0.0002 + 0.0001}{0.0006}\)
\(\frac{1}{C_T}\) = \(\frac{0.0003}{0.0006}\)
C\(_T\) = 0.0002\muF.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70682)
Physics
WAEC / WASSCE 2024
Electric Field
Solution
Two isolated charged spheres each of magnitude + 1.0 x 10\(^{-6}\) C are seperated by a distance of 2.0 x 10\(^{-1}\) m. Calculate the magnitude of the electrostatic force between them [K = \(\frac{1}{4\pi \epsilon_o}\) = 9 x 10\(^9\)Nm\(^2\)C\(^{-2}\)]
A
2.25 x 10\(^{-1}\)N
B
4.50 x 10\(^{-1}\)N
C
2.25 x 10\(^1\)N
D
4.50 x 10\(^1\)N
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
F = \(\frac{Kq_1q_2}{r^2}\)
F = \(\frac{9 \times 10^9 \times (1.0 \times 10^{-6})^2}{(0.2)^2}\)
= 2.25 x 10\(^{-1}\)N
Source: WAEC / JAMB / NECO Classroom Archive (Record #72047)
Physics
WAEC / WASSCE 2024
Electric Field
Diagram
Solution
In the circuit diagram above, which lamp causes the other to go off?
Attached Figure / Diagram
A
P
B
Q
C
R
D
S
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
If lamp S goes off, all other lamps will go off. This is because lamp S is required to complete the whole circuit.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72049)
Physics
WAEC / WASSCE 2024
Electric Field
Solution
A capacitor described as a mica capacitor has
A
both plates are made of mica.
B
a mica dielectric.
C
a dielectric made of mica
D
the same capacitance as that of mica
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A mica capacitor has mica as its dielectric material. Known for excellent stability, low loss, and high insulation resistance, they are used in precision applications, particularly in RF circuits and high-frequency settings.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72064)
Physics
UTME / JAMB 2024
Current Electricity
Solution
Electrolysis can be investigated using
A
Voltmeter
B
Ammeter
C
Galvanometer
D
Voltameter
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
A voltameter measures the amount of substance that is deposited during electrolysis, allowing for the study of the electrochemical process. It provides information about the quantity of electric charge passed through the electrolyte and the resulting chemical changes. Voltameter is also called an Electrolytic cell.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70367)
Physics
UTME / JAMB 2024
Current Electricity
Solution
If a sheet of an insulating material is inserted between the plates of an air capacitor, the capacitance will
A
become zero
B
decrease
C
increase
D
remain unchanged
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
If a sheet of an insulating material (dielectric) is inserted between the plates of an air capacitor, the capacitance will increase.
When a dielectric material is inserted: The permittivity \(\epsilon\) increases (since the permittivity of the dielectric is greater than that of air),
This results in an increase in capacitance.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70373)