Question 35 of 276

Two circular coils $P$ and $Q$ of $100$ turns each have same radius of $\pi cm$. 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 $\sqrt{x}mT$, where $x=$ ______. [Use ${\mu }_{0}=4\pi \times {10}^{-7}Tm{A}^{-1}$]

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Answer: 20

Consider two coils as shown below:

Magnetic field due to P, ${B}_{P}=N\left(\dfrac{{\mu }_{0}{i}_{1}}{2r}\right)=100\times \dfrac{{\mu }_{0}\times 1}{2\left(\pi \times {10}^{-2}\right)}=2\times {10}^{-3}T$
Magnetic field due to Q,
${B}_{Q}=N\left(\dfrac{{\mu }_{0}{i}_{2}}{2r}\right)=100\times \dfrac{{\mu }_{0}\times 2}{2\pi \times {10}^{-2}}=4\times {10}^{-3}T$
Therefore, ${B}_{net}=\sqrt{{B}_{P}^{2}+{B}_{Q}^{2}}$
$=\sqrt{20}mT$
Hence, $x=20$