NTA Abhyas JEE Main2020PhysicsElectromagnetic InductionPractice
A uniform but time-varying magnetic field is present in a circular region of radius R . The magnetic field is perpendicular and into the plane of the paper and the magnitude of the field is increasing at a constant rate α . There is a straight conducting rod of length 2 R placed as shown in the diagram. The magnitude of induced E.M.F. across the rod is
Options
- Aπ R 2 α
- Bπ R 2 α 4
- CR 2 α 2
- Dπ R 2 α 2
Correct answer
B. π R 2 α 4
Step-by-step solution
Considering a square loop as shown, E.M.F. across the loop is given by e = ∮ E.dl = - d ϕ dt = - A dB dt (where A = area = π R 2 ) ⇒    e = - π R 2 dB dt = - π R 2 α ∴ Magnitude of E.M.F. across one side = π R 2 α 4