JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 24)
Two circular coils $$P$$ and $$Q$$ of 100 turns each have same radius of $$\pi \mathrm{~cm}$$. 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 $$\sqrt{x} ~m T$$, where $$x=$$ __________.
[Use $$\mu_0=4 \pi \times 10^{-7} \mathrm{~TmA}^{-1}$$]
Explanation
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$$\begin{aligned} & \mathrm{B}_{\mathrm{P}}=\frac{\mu_0 \mathrm{Ni}_1}{2 \mathrm{r}}=\frac{\mu_0 \times 1 \times 100}{2 \pi}=2 \times 10^{-3} \mathrm{~T} \\ & \mathrm{~B}_{\mathrm{Q}}=\frac{\mu_0 \mathrm{Ni}_2}{2 \mathrm{r}}=\frac{\mu_0 \times 2 \times 100}{2 \pi}=4 \times 10^{-3} \mathrm{~T} \\ & \mathrm{~B}_{\text {net }}=\sqrt{\mathrm{B}_{\mathrm{P}}^2+\mathrm{B}_{\mathrm{Q}}^2} \\ & =\sqrt{20} \mathrm{mT} \\ & \mathrm{x}=20 \end{aligned}$$
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