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Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 31 January, Shift 2 — Question 53

Two circular coils PP and QQ of 100 turns each have same radius of πcm\pi \mathrm{cm}. The currents in P and R are 1 A and 2 A 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\sqrt{\mathrm{x}} \mathrm{mT}, where x=\mathrm{x}= _______\_\_\_\_\_\_\_. [\left[\right. Use μ0=4π×10−7TmA−1]\left.\mu_{0}=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}\right]

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

BP=μ0Ni12r=μ0×1×1002π=2×10−3 T\mathrm{B}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{Ni}_{1}}{2 \mathrm{r}}=\frac{\mu_{0} \times 1 \times 100}{2 \pi}=2 \times 10^{-3} \mathrm{~T}

BQ=μ0Ni22r=μ0×2×1002π=4×10−3 T\mathrm{B}_{\mathrm{Q}}=\frac{\mu_{0} \mathrm{Ni}_{2}}{2 \mathrm{r}}=\frac{\mu_{0} \times 2 \times 100}{2 \pi}=4 \times 10^{-3} \mathrm{~T}

Bnet=BP2+BQ2\mathrm{B}_{\mathrm{net}}=\sqrt{\mathrm{B}_{\mathrm{P}}^{2}+\mathrm{B}_{\mathrm{Q}}^{2}}

=20mT=\sqrt{20} \mathrm{mT} x=20\mathrm{x}=20

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law

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