Two circular coils P and Q of 100 turns same radius

JEE MAIN 2024

31.01.24 S2

Question

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 $\sqrt{\mathrm{x}} \mathrm{mT}$, where $\mathrm{x}=$ $\_\_\_\_$ . [Use $\mu_0=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}$ ]

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Solution

$\mathrm{B}_{\mathrm{P}}=\frac{\mu_0 \mathrm{Ni}_1}{2 \mathrm{r}}=\frac{\mu_0 \times 1 \times 100}{2 \pi}=2 \times 10^{-3} \mathrm{~T}$ $\begin{aligned} & \mathrm{B}_{\mathrm{Q}}=\frac{\mu_0 \mathrm{Ni}_2}{2 \mathrm{r}}=\frac{\mu_0 \times 2 \times 100}{2 \pi}=4 \times 10^{-3} \mathrm{~T} \\ & \mathrm{~B}_{\text {net }}=\sqrt{\mathrm{B}_{\mathrm{P}}^2+\mathrm{B}_{\mathrm{Q}}^2} \\ & =\sqrt{20} \mathrm{mT} \\ & \mathrm{x}=20\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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