Q. A capacitor in an AC circuit has a capacitive reactance of 50 ohms. If the frequency of the AC source is increased, what happens to the capacitive reactance?
A.Increases
B.Decreases
C.Remains the same
D.Becomes infinite
Solution
Capacitive reactance (X_C) is given by X_C = 1/(2πfC). If the frequency (f) increases, X_C decreases.
Q. A capacitor in an AC circuit has a capacitive reactance of 50 ohms. What is the frequency if the capacitance is 10 microfarads?
A.1 kHz
B.10 kHz
C.100 Hz
D.1000 Hz
Solution
Capacitive reactance (X_C) is given by X_C = 1 / (2πfC). Rearranging gives f = 1 / (2πX_CC). Substituting X_C = 50 ohms and C = 10 x 10^-6 F gives f = 318.31 Hz, approximately 1 kHz.
Q. A charged particle moves in a magnetic field B with a velocity v. What is the expression for the magnetic force acting on the particle?
A.qvB
B.qvBsinθ
C.qvBcosθ
D.qB
Solution
The magnetic force acting on a charged particle moving in a magnetic field is given by F = qvBsinθ, where θ is the angle between the velocity and the magnetic field.
Q. A charged particle moves in a magnetic field. What is the condition for the particle to experience maximum force?
A.Velocity is zero
B.Velocity is parallel to the field
C.Velocity is perpendicular to the field
D.Charge is zero
Solution
The magnetic force on a charged particle is given by F = qvB sin(θ). The force is maximum when the angle θ is 90 degrees, meaning the velocity is perpendicular to the magnetic field.
Correct Answer: C — Velocity is perpendicular to the field
Q. A charged particle moves in a magnetic field. What is the condition for the particle to experience no magnetic force?
A.The particle must be at rest
B.The particle must be moving parallel to the magnetic field
C.The particle must be moving perpendicular to the magnetic field
D.The magnetic field must be zero
Solution
The magnetic force on a charged particle is given by F = q(v × B). If the velocity vector v is parallel to the magnetic field B, the cross product is zero, resulting in no magnetic force.
Correct Answer: B — The particle must be moving parallel to the magnetic field
Q. A charged particle moves in a magnetic field. What is the effect of the magnetic field on the particle's motion?
A.It accelerates the particle
B.It changes the particle's speed
C.It changes the particle's direction
D.It has no effect
Solution
A magnetic field exerts a force on a charged particle that is perpendicular to both the velocity of the particle and the magnetic field, changing its direction but not its speed.
Correct Answer: C — It changes the particle's direction
Q. A charged particle moves in a magnetic field. What is the nature of the force acting on it?
A.Always in the direction of motion
B.Always opposite to the direction of motion
C.Perpendicular to the direction of motion
D.Depends on the charge of the particle
Solution
The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force law, which states that the force is perpendicular to both the velocity of the particle and the magnetic field.
Correct Answer: C — Perpendicular to the direction of motion
Q. A circular loop is placed in a uniform magnetic field. If the loop is rotated about its diameter, what happens to the induced EMF?
A.It increases
B.It decreases
C.It remains constant
D.It becomes zero
Solution
When the loop is rotated about its diameter, the angle between the magnetic field and the normal to the loop changes, but the magnetic flux remains constant. Therefore, the induced EMF becomes zero as there is no change in magnetic flux.
Q. A circular loop of radius R carries a current I. What is the magnetic field at the center of the loop?
A.μ₀I/(2R)
B.μ₀I/R
C.μ₀I/(4R)
D.μ₀I/(8R)
Solution
The magnetic field at the center of a circular loop carrying current I is given by the formula B = (μ₀I)/(2R), where μ₀ is the permeability of free space.
Q. A circular loop of radius r is placed in a uniform magnetic field B. If the magnetic field is perpendicular to the plane of the loop, what is the magnetic flux through the loop?
A.0
B.πr²B
C.2πrB
D.B/r
Solution
The magnetic flux Φ through a surface is given by Φ = B * A, where A is the area. For a circular loop, A = πr², so Φ = B * πr².
Q. A circular loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the magnetic field strength is doubled?
A.Induced EMF is halved
B.Induced EMF remains the same
C.Induced EMF is doubled
D.Induced EMF is quadrupled
Solution
According to Faraday's law of electromagnetic induction, the induced EMF is directly proportional to the rate of change of magnetic flux. If the magnetic field strength is doubled, the induced EMF will also double.
Q. A circular loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the area of the loop is increased?
A.Increases
B.Decreases
C.Remains the same
D.Depends on the magnetic field strength
Solution
According to Faraday's law of electromagnetic induction, the induced EMF is proportional to the rate of change of magnetic flux. Increasing the area increases the flux, thus increasing the induced EMF.
Q. A coil of wire is placed in a magnetic field. If the magnetic field strength is increased, what happens to the induced EMF in the coil?
A.It increases
B.It decreases
C.It remains the same
D.It becomes zero
Solution
According to Faraday's law of electromagnetic induction, the induced EMF in a coil is directly proportional to the rate of change of magnetic flux. Increasing the magnetic field strength increases the magnetic flux, thus increasing the induced EMF.
