Physics · Electromagnetic Induction
JEE Main 2024 — 27 January, Shift 1 — Question 39
A rectangular loop of length 2.5 m and width 2 m is placed at to a magnetic field of 4 T . The loop is removed from the field in 10 sec . The average emf induced in the loop during this time is
- Option A:
-2 V
- Option B:
- Option C:Correct
+1 V
- Option D:
-1 V
Answer: C
Step-by-step solution
EMF =-( Change in magnetic flux / Time) Magnetic field (B) = 4 T Length = 2.5 m Width = 2 m Area (A) = length × width = 2.5 × 2 = 5 m² Angle with magnetic field = 60° Time (t) = 10 s Initial flux = B × A × cos(60°) = 4 × 5 × 0.5 = 10 Weber
Final flux = 0 (since loop is removed from the field)
Change in flux = 0-10 = -10 Weber Average EMF = -Change in flux / Time = 10 / 10 = 1 volt
Final Answer: 1 V
Answer key and solution verified before publishing.
Practise Electromagnetic Induction
Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.
- Exam
- JEE Main 2024
- Paper
- 27 January, Shift 1
- Subject
- Physics
- Chapter
- Electromagnetic Induction
- Topic
- Magnetic Flux, Faraday's Law and Lenz's Law