| ⇦ | 
| ⇨ |
A conductor of length 5 cm is moved parallel to itself with a speed of 2 m/s, perpendicular to a uniform magnetic field of 10⁻³ Wb/m². The induced emf generated is
Options
(a) 2×10⁻³ V
(b) 1×10⁻³ V
(c) 1×10⁻⁴ V
(d) 2×10⁻⁴ V
Correct Answer:
1×10⁻⁴ V
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - What determines the nature of the path followed by the particle?
- Two bodies of masses m₁ and m₂ are initially at rest at infinite distance apart
- A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support.
- A particle doing SHM has amplitude=4 cm and time period = 12 sec. The ratio between times
- The electric current in AC circuit is given by the relation i-3 sinωt+4 cos ωt.
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- What determines the nature of the path followed by the particle?
- Two bodies of masses m₁ and m₂ are initially at rest at infinite distance apart
- A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support.
- A particle doing SHM has amplitude=4 cm and time period = 12 sec. The ratio between times
- The electric current in AC circuit is given by the relation i-3 sinωt+4 cos ωt.
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply