physics : class12 Rotating Conductors : Induced Current : Electromagnetic Induction and Alternating Currents MCQ Questions and Answers
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Rotating Conductors
MCQ questions and answers.
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MCQ Questions
1. A conducting rod of length L rotates with constant angular speed ω about one of its ends in a uniform magnetic field B perpendicular to the plane of rotation. What is the induced emf between the axis and the free end?
A.
BωL²
B.
½BωL²
C.
½Bω²L
D.
BωL
Correct Answer:
½BωL²
Explanation:
An element at distance r from the axis has speed ωr, so its motional emf is dε = Bωr dr. Integrating from r = 0 to L gives ε = ∫₀ᴸ Bωr dr = ½BωL².
2. A conducting rod of length 2L rotates about an axis through its midpoint, perpendicular to the rod and to a uniform magnetic field B. If the rod rotates with angular speed ω, what is the potential difference between its two ends?
A.
BωL²
B.
2BωL²
C.
Zero
D.
½BωL²
Correct Answer:
Zero
Explanation:
The motional emf from the midpoint to either end has magnitude ½BωL², but the two halves produce equal and opposite potential changes. Therefore the net potential difference between the two ends is zero.
3. A conducting disc of radius R rotates with angular speed ω about its symmetry axis in a uniform magnetic field B parallel to the axis. The resistance between the centre and rim is r. If a resistor of resistance R₀ is connected between the centre and rim, what is the current through the external resistor?
A.
BωR²/(r + R₀)
B.
BωR²/[2(r + R₀)]
C.
BωR/(r + R₀)
D.
BωR²/[2rR₀]
Correct Answer:
BωR²/[2(r + R₀)]
Explanation:
The rotating disc acts as a homopolar generator with emf ε = ½BωR² between its centre and rim. The total circuit resistance is r + R₀, so I = ε/(r + R₀) = BωR²/[2(r + R₀)].
4. A radial conducting rod of length L and resistance r rotates about one end in a magnetic field B perpendicular to its plane of rotation. Its free end is connected to the axis through an external resistance R. What mechanical torque must be supplied to maintain angular speed ω, neglecting friction?
A.
B²ωL⁴/[4(r + R)]
B.
B²ωL²/[2(r + R)]
C.
BωL²/[2(r + R)]
D.
B²ω²L⁴/[4(r + R)]
Correct Answer:
B²ωL⁴/[4(r + R)]
Explanation:
The induced emf is ε = ½BωL² and the current is I = ε/(r + R). The opposing magnetic torque satisfies τω = Iε, hence τ = ε²/[ω(r + R)] = B²ωL⁴/[4(r + R)].
5. A conducting rod of length L rotates about one end with angular velocity ω in a uniform magnetic field whose component perpendicular to the plane of rotation is B cosθ. What is the induced emf between the axis and the free end?
A.
BωL² cosθ
B.
½BωL² cosθ
C.
½BωL² sinθ
D.
BωL cosθ
Correct Answer:
½BωL² cosθ
Explanation:
Only the magnetic-field component normal to the plane swept by the rod contributes to the motional emf. Replacing B by B cosθ in ε = ½BωL² gives ε = ½BωL² cosθ.
6. A conducting radial rod rotates in a uniform magnetic field B with angular speed ω. The rod is connected between radii a and b, where b > a, and its resistance is r. What emf is generated between these two radii?
A.
½Bω(b² − a²)
B.
Bω(b − a)²
C.
½Bω(b − a)²
D.
Bω(b² − a²)
Correct Answer:
½Bω(b² − a²)
Explanation:
For an element at radius x, dε = Bωx dx. Integrating between a and b gives ε = ∫ₐᵇ Bωx dx = ½Bω(b² − a²).
7. A rotating conducting rod of length L has a resistance r and generates an emf ε = ½BωL². It is connected to a load resistance R. What fraction of the mechanical power supplied to maintain constant rotation is delivered to the load?
A.
r/(R + r)
B.
R/(R + r)
C.
R²/(R + r)²
D.
rR/(R + r)²
Correct Answer:
R/(R + r)
Explanation:
The current is I = ε/(R + r). Mechanical input power equals total electrical dissipation, I²(R + r), while load power is I²R. Their ratio is P_load/P_mechanical = R/(R + r).
8. A conducting disc rotates clockwise as viewed from above in a uniform magnetic field directed upward. What is the direction of conventional current in an external circuit connected between the centre and the rim, assuming the circuit is closed?
A.
From centre to rim through the external circuit
B.
From rim to centre through the external circuit
C.
No current flows because the disc is rotating uniformly
D.
The current reverses direction every half rotation
Correct Answer:
From centre to rim through the external circuit
Explanation:
For a positive charge at radius r, v is tangential and B is upward. The magnetic force q(v × B) is radially outward for clockwise rotation, making the rim positive relative to the centre. Conventional current therefore flows from rim to centre inside the disc and from centre to rim through the external circuit.
9. A conducting rod of length L rotates about one end in a uniform magnetic field B perpendicular to its plane of rotation. If the angular speed is doubled while the rod length is halved, how does the induced emf change?
A.
It becomes four times larger
B.
It becomes twice as large
C.
It becomes half as large
D.
It remains unchanged
Correct Answer:
It becomes half as large
Explanation:
The emf is ε = ½BωL². With ω′ = 2ω and L′ = L/2, ε′/ε = [2ω(L/2)²]/(ωL²) = 1/2. Thus the emf becomes half its original value.
10. A conducting disc of radius R rotates in a uniform axial magnetic field B. The angular speed varies with time as ω = αt, where α is constant and the disc is connected to a resistance Rₑ between its centre and rim. Neglecting the disc's internal resistance, what is the current at time t?
A.
BαR²/(Rₑ)
B.
BαtR²/(2Rₑ)
C.
Bαt²R²/(2Rₑ)
D.
BαR/(2Rₑ)
Correct Answer:
BαtR²/(2Rₑ)
Explanation:
At any instant, the rotating disc produces emf ε = ½BωR². Since ω = αt, ε = ½BαtR². Dividing by the external resistance gives I = BαtR²/(2Rₑ).
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