Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A magnet of moment 4 m2 is kept suspended in a magnetic field of induction 5×10-5 T. The work done in rotating it through 1800 is
PhysicsMagnetic Properties of MatterNEET
Options:
  • A 4×10-4 J
  • B 5×10-4 J
  • C 2×10-4 J
  • D 10-4 J
Solution:
1596 Upvotes Verified Answer
The correct answer is: 4×10-4 J
W=M.B(cos θ f cos θ i )

=( )4×5× 10 5 (cos180cos0)

=20× 10 5 ( 11 )

=4×104 J

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.