Exams › JEE Main › Physics
Two neighbouring coils have a mutual inductance of 1.5 H. If the current in one coil rises from 0 to 20 A in 0.5 s, what is the change in the flux linkage of the other coil?
- 30 Wb
- 15 Wb
- 60 Wb
- 10 Wb
Correct answer: 30 Wb
Solution
The flux linkage of the second coil is N*phi = M*I. The change is delta(N*phi) = M*delta(I) = 1.5 * (20 - 0) = 30 Wb. The 0.5 s interval is needed only if the induced EMF were asked.
Related JEE Main Physics questions
- A metallic ring is kept in a horizontal position, and a bar magnet is released so that it falls through the ring with its length aligned along the ring’s axis. The magnet’s acceleration while falling is
- A coil has resistance 10 Ω, and the magnetic flux linked with it changes with time according to ϕ = 4t² + 2t + 1 (in weber). The current induced in the coil at t = 1 s is
- Two coils are kept near one another. The mutual inductance of the coil pair depends on
- A conducting rod of length ℓ is attached to a string of length 2ℓ and is rotated on a horizontal table about the fixed end of the string with angular speed ω. If a uniform magnetic field B acts vertically through the region, what is the induced emf between the two ends of the rod?
- A pair of coils, each having N turns, has mutual inductance M henry. If the current in one coil changes from 1 ampere to zero in time t seconds, what is the induced emf per turn in the other coil, in volts?
- A boat travels straight toward the east in a place where Earth’s magnetic field is horizontal, points due north, and has magnitude 5.0 × 10⁻⁵ N A⁻¹ m⁻¹. A 2 m long vertical antenna is mounted on the boat. If the boat’s speed is 1.50 m s⁻¹, what emf is induced across the antenna wire?
⚔️ Practice JEE Main Physics free + battle 1v1 →