JEE Main 2023 Physics question on Electromagnetic Induction and AC
A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be ___________
A.
16 V
B.
8 V
C.
12 V
D.
20 V
Show answer and solution
Correct option: A
Solution
Using the motional EMF formula: emf = BLv = 2 × 1 × 8 = 16 V.
[JEE Main 2023 · 25 Jan · Shift 2]
Practise Electromagnetic Induction and AC the way it is asked
A timed set of real JEE Main Electromagnetic Induction and AC questions, graded as the exam grades. No sign-up.
