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 moving coil galvanometer has 100 turns and each turn has an area of 2.0 cm2. The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is x×10-5 N-m/rad. The value of x is ______ .
PhysicsMagnetic Effects of CurrentJEE MainJEE Main 2024 (01 Feb Shift 2)
Solution:
1433 Upvotes Verified Answer
The correct answer is: 4

Torque produced in a moving coil galvanometer is, τ=BINAsinϕ.

Let torsional constant of the suspension wire be C, then Cθ=BINAsin90°

C= BINA θ=0.01×10×10-3×100×2×10-40.05

=4×10-5 N-m/rad.

Therefore, x=4.

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.