Motional Emf
A straight conducting rod of length 0.5 m slides at a constant speed of 4 m/s along frictionless rails in a uniform magnetic field of 0.3 T directed perpendicular to the plane of motion. What EMF appears across the ends of the rod?
Select the correct option:
Solution
0.6 V
Motional EMF arises because the free charges inside a conductor moving through a magnetic field experience a magnetic force (q\vec{v}\times\vec{B}) that drives them along the rod until an electrostatic field balances it, establishing a steady potential difference. When the rod, its velocity, and the field are mutually perpendicular, the induced EMF takes the simple form (\varepsilon = BvL). Inserting (B = 0.3) T, (v = 4) m/s and (L = 0.5) m yields (\varepsilon = 0.3 \times 4 \times 0.5 = 0.6) V. The option 0.3 V wrongly omits the length factor. The option 1.2 V doubles the result by mistakenly using twice the rod length. The option 0.15 V halves the speed contribution. This is exactly the rail-and-rod arrangement treated in NCERT to introduce motional EMF as an alternative route to Faraday's law. A unit and magnitude check confirms T·(m/s)·m gives volts, and the small value is consistent with a short rod moving slowly in a modest field.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- motional emf
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
0.6 V
Motional EMF arises because the free charges inside a conductor moving through a magnetic field experience a magnetic force (q\vec{v}\times\vec{B}) that drives them along the rod until an electrostatic field balances it, establishing a steady potential difference. When the rod, its velocity, and the field are mutually perpendicular, the induced EMF takes the simple form (\varepsilon = BvL). Inserting (B = 0.3) T, (v = 4) m/s and (L = 0.5) m yields (\varepsilon = 0.3 \times 4 \times 0.5 = 0.6) V. The option 0.3 V wrongly omits the length factor. The option 1.2 V doubles the result by mistakenly using twice the rod length. The option 0.15 V halves the speed contribution. This is exactly the rail-and-rod arrangement treated in NCERT to introduce motional EMF as an alternative route to Faraday's law. A unit and magnitude check confirms T·(m/s)·m gives volts, and the small value is consistent with a short rod moving slowly in a modest field.
This easy difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of motional emf. It appeared in the 2025 exam.
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