Consider two coils as shown below:

Magnetic field due to P, BP=N(2rμ0i1)=100×2(π×10−2)μ0×1=2×10−3T
Magnetic field due to Q,
BQ=N(2rμ0i2)=100×2π×10−2μ0×2=4×10−3T
Therefore, Bnet=BP2+BQ2
=20mT
Hence, x=20
JEE Main 2024 — Physics Electromagnetism
Two circular coils P and Q of 100 turns each have same radius of πcm. The currents in P and R are 1A and 2A respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is xmT, where x= ______. [Use μ0=4π×10−7TmA−1]
Held on 31 Jan 2024 · Verified 6 Jul 2026.
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