Exams › JEE Main › Physics
Two coaxial solenoids are wound with thin insulated wire on the same pipe of cross-sectional area A = 10 cm² and length 20 cm. One solenoid has 300 turns and the other has 400 turns. Find their mutual inductance. (mu0 = 4*pi*10⁻⁷ T m/A)
- 2.4*pi*10⁻⁵ H
- 4.8*pi*10⁻⁴ H
- 4.8*pi*10⁻⁵ H
- 2.4*pi*10⁻⁴ H
Correct answer: 2.4*pi*10⁻⁵ H
Solution
The mutual inductance of two coaxial solenoids sharing the same cross-section is M = mu0*N1*N2*A/l. Substituting: M = (4*pi*10⁻⁷)*(300)*(400)*(10*10⁻⁴)/(0.20). Compute the numerator coefficient: 4*pi*10⁻⁷ * 1.2*10⁵ * 1.0*10⁻³ = 4*pi*10⁻⁷ * 1.2*10² = 4.8*pi*10⁻⁵; dividing by 0.20 gives 2.4*pi*10⁻⁵ H.
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 →