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 coil of 100 turns and area 5 square centimetre is placed in a magnetic field $B=0.2 T$. The normal to the plane of the coil makes an angle of $60^{\circ}$ with the direction of the magnetic field. The magnetic flux linked with the coil is
PhysicsElectromagnetic InductionJEE Main
Options:
  • A ${ }^{5 \times 10^{-3} \mathrm{~Wb}}$
  • B ${ }^{5 \times 10^{-5} \mathrm{~Wb}}$
  • C $0^{10^{-2} \mathrm{~Wb}}$
  • D $10^{-4} \mathrm{~Wb}$
Solution:
1691 Upvotes Verified Answer
The correct answer is: ${ }^{5 \times 10^{-3} \mathrm{~Wb}}$
$\phi=N B A \cos \theta=100 \times 0.2 \times 5 \times 10^{-4} \cos 60^{\circ}$
$=5 \times 10^{-3} \mathrm{~Wb}$

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.