A small piece of metal wire is dragged across the gap between the poles of a magnet in…
Physics · JEE Main · NTA Exams — Electromagnetic Induction and Alternating Currents
A small piece of metal wire is dragged across the gap between the poles of a magnet in 0.4 s. If change in magnetic flux in the wire is \(8 \times 10^{-4}\) Wb, then emf induced in the wire is
- \(8 \times 10^{-3} \mathrm{~V}\)
- \(6 \times 10^{-3} \mathrm{~V}\)
- \(4 \times 10^{-3} \mathrm{~V}\)
- \(2 \times 10^{-3} \mathrm{~V}\)
Answer
(A) 8 10^ -3 ~V
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- A straight conductor of length 0.4 m is moved with a speed of 7 ms –1 perpendicular to a magnetic field of…
- The potential difference V cd across the inductor L is
- The coil of a dynamo has N turns, area A and it rotates with an angular velocity ω in a uniform magnetic…
- The average power dissipation in a pure capacitor in AC circuit is
- The inductance between A and D is
- A cylindrical conducting rod is kept with its axis along the x - axis. Also there exists a uniform magnetic…
- If a coil of 40 turns and area 4.0 cm 2 is suddenly removed from a magnetic field, it is observed that a…
- Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area A = 10 cm 2…
More Electromagnetic Induction and Alternating Currents questions · Browse all practice questions