Motional EMF — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Motional EMF MCQs with step-by-step solutions (21 questions). Part of Electromagnetic Induction. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Motional EMF · easy · theory
A straight conductor of length $l$ moves with velocity $v$ perpendicular to a field $B$. The EMF induced across it is:
A. $\dfrac{Bv}{l}$
B. $Blv^2$
C. $Blv$ ✓ Correct
D. $\dfrac{Bl}{v}$
Solution: In one second the rod sweeps an area $lv$, so the flux cut per second is $Blv$.
Q2 — Motional EMF · medium · theory
Motional EMF arises because the free charges in the moving conductor experience:
A. An electrostatic force from the field
B. A magnetic force $qvB$ along the length of the conductor ✓ Correct
C. No force at all
D. A gravitational force
Solution: The magnetic force drives charges to the ends until the electric field they set up balances it.
Q3 — Motional EMF · easy · theory
The direction of the motional EMF in a moving conductor is found using:
A. Fleming's right-hand rule ✓ Correct
B. Fleming's left-hand rule
C. The right-hand thumb rule
D. Lenz's law only
Solution: With the forefinger along the field and the thumb along the motion, the middle finger gives the induced current.
Q4 — Motional EMF · medium · theory
A rod of length $l$ rotating about one end with angular velocity $\omega$ in a field $B$ perpendicular to its plane develops an EMF of:
A. $B\omega l^2$
B. $\dfrac{1}{2}B\omega l^2$ ✓ Correct
C. $2B\omega l^2$
D. $\dfrac{1}{2}B\omega l$
Solution: Each element sweeps a different area; integrating along the rod gives the factor of one-half.
Q5 — Motional EMF · medium · theory
A metal disc of radius $R$ rotating with angular velocity $\omega$ about its axis in a field $B$ parallel to the axis develops between centre and rim an EMF of:
A. Zero
B. $\dfrac{1}{2}B\omega R$
C. $\dfrac{1}{2}B\omega R^2$ ✓ Correct
D. $B\omega R^2$
Solution: Each radius behaves like a rotating rod, giving the same one-half factor.
Q6 — Motional EMF · medium · theory
When a rod moves in a magnetic field with a closed circuit, the mechanical power supplied equals:
A. The electrical power dissipated in the circuit ✓ Correct
B. Half the electrical power dissipated
C. Zero
D. The magnetic energy stored
Solution: Energy conservation requires the work done against the opposing magnetic force to reappear as electrical energy.
Q7 — Motional EMF · hard · theory
An aircraft flying horizontally develops an EMF between its wing tips because of:
A. The vertical component of the Earth's magnetic field ✓ Correct
B. The horizontal component of the Earth's magnetic field
C. Friction with the air
D. The electric field of the atmosphere
Solution: Only the field component perpendicular to both the wingspan and the velocity contributes.
Q8 — Motional EMF · medium · theory
The motional EMF developed across a moving rod is:
A. Zero in an open circuit
B. Proportional to the resistance of the circuit
C. Inversely proportional to the resistance
D. Independent of the resistance of the circuit ✓ Correct
Solution: The EMF is fixed by $Blv$; the resistance determines only the current that results.
Q9 — Motional EMF · easy · numerical
A rod of length $0.2\text{ m}$ moves at $10\text{ m/s}$ perpendicular to a field of $0.5\text{ T}$. The EMF induced is:
A. $10\text{ V}$
B. $0.1\text{ V}$
C. $1\text{ V}$ ✓ Correct
D. $0.25\text{ V}$
Solution: $e = Blv = 0.5 \times 0.2 \times 10 = 1\text{ V}$.
Q10 — Motional EMF · easy · numerical
A rod of length $1\text{ m}$ moves at $5\text{ m/s}$ perpendicular to a field of $0.2\text{ T}$. The EMF induced is:
A. $5\text{ V}$
B. $0.2\text{ V}$
C. $1\text{ V}$ ✓ Correct
D. $10\text{ V}$
Solution: $e = Blv = 0.2 \times 1 \times 5 = 1\text{ V}$.
Q11 — Motional EMF · medium · numerical
A rod of length $0.5\text{ m}$ moves at $4\text{ m/s}$ perpendicular to a field of $0.3\text{ T}$. The EMF induced is:
A. $0.6\text{ V}$ ✓ Correct
B. $0.06\text{ V}$
C. $1.2\text{ V}$
D. $6\text{ V}$
Solution: $e = 0.3 \times 0.5 \times 4 = 0.6\text{ V}$.
Q12 — Motional EMF · hard · numerical
A copper disc of radius $0.1\text{ m}$ rotates at $10\text{ rev/s}$ about its axis in a field of $0.2\text{ T}$ parallel to the axis. The EMF between centre and rim is approximately:
A. $0.63\text{ V}$
B. $0.126\text{ V}$
C. $0.063\text{ V}$ ✓ Correct
D. $0.0126\text{ V}$
Solution: $e = \dfrac{1}{2}B\omega R^2 = \dfrac{1}{2}(0.2)(2\pi \times 10)(0.01) \approx 0.063\text{ V}$.
Q13 — Motional EMF · hard · numerical
A rod of length $1\text{ m}$ rotates about one end at $10\text{ rad/s}$ in a field of $0.5\text{ T}$ perpendicular to its plane of rotation. The EMF is:
A. $5\text{ V}$
B. $1.25\text{ V}$
C. $10\text{ V}$
D. $2.5\text{ V}$ ✓ Correct
Solution: $e = \dfrac{1}{2}B\omega l^2 = \dfrac{1}{2}(0.5)(10)(1) = 2.5\text{ V}$.
Q14 — Motional EMF · easy · numerical
If the speed of a rod moving in a magnetic field is doubled, the motional EMF:
A. Remains unchanged
B. Halves
C. Doubles ✓ Correct
D. Becomes four times
Solution: $e = Blv \propto v$.
Q15 — Motional EMF · easy · numerical
A rod develops an EMF of $2\text{ V}$ and is connected to a circuit of total resistance $4\,\Omega$. The induced current is:
A. $2\text{ A}$
B. $8\text{ A}$
C. $0.5\text{ A}$ ✓ Correct
D. $0.25\text{ A}$
Solution: $I = \dfrac{e}{R} = \dfrac{2}{4} = 0.5\text{ A}$.
Q16 — Motional EMF · medium · numerical
A rod develops an EMF of $1\text{ V}$ in a circuit of resistance $0.5\,\Omega$. The power dissipated is:
A. $0.5\text{ W}$
B. $2\text{ W}$ ✓ Correct
C. $1\text{ W}$
D. $4\text{ W}$
Solution: $P = \dfrac{e^2}{R} = \dfrac{1}{0.5} = 2\text{ W}$.
Q17 — Motional EMF · hard · numerical
An aircraft of wingspan $30\text{ m}$ flies at $300\text{ m/s}$ where the vertical component of the Earth's field is $5 \times 10^{-5}\text{ T}$. The EMF between the wing tips is:
A. $1.5\text{ V}$
B. $0.045\text{ V}$
C. $4.5\text{ V}$
D. $0.45\text{ V}$ ✓ Correct
Solution: $e = B_Vlv = 5 \times 10^{-5} \times 30 \times 300 = 0.45\text{ V}$.
Q18 — Motional EMF · medium · numerical
A rod of length $0.4\text{ m}$ moves at $8\text{ m/s}$ perpendicular to a field of $0.25\text{ T}$. The EMF is:
A. $8\text{ V}$
B. $0.8\text{ V}$ ✓ Correct
C. $0.08\text{ V}$
D. $1.6\text{ V}$
Solution: $e = 0.25 \times 0.4 \times 8 = 0.8\text{ V}$.
Q19 — Motional EMF · medium · numerical
A rod of length $0.6\text{ m}$ in a field of $0.5\text{ T}$ develops an EMF of $3\text{ V}$. Its speed is:
A. $10\text{ m/s}$ ✓ Correct
B. $100\text{ m/s}$
C. $0.9\text{ m/s}$
D. $1\text{ m/s}$
Solution: $v = \dfrac{e}{Bl} = \dfrac{3}{0.5 \times 0.6} = \dfrac{3}{0.3} = 10\text{ m/s}$.
Q20 — Motional EMF · medium · numerical
If the radius of a rotating disc is doubled at the same angular velocity and field, the EMF between centre and rim becomes:
A. Twice as large
B. Four times as large ✓ Correct
C. Unchanged
D. Half as large
Solution: $e = \dfrac{1}{2}B\omega R^2 \propto R^2$.
Q21 — Motional EMF · hard · numerical
A rod moves with velocity $v$ making an angle $\theta$ with a magnetic field $B$. The EMF induced across its length $l$ is:
A. $Blv\cos\theta$
B. $Blv$
C. $Blv\tan\theta$
D. $Blv\sin\theta$ ✓ Correct
Solution: Only the velocity component perpendicular to the field sweeps flux, giving the sine factor.