Magnetic Effects Of Current Question 15
Question 15 - 2024 (31 Jan Shift 2)
Two circular coils P and Q of 100 turns each have same radius of $\pi \mathrm{cm}$. The currents in $\mathrm{P}$ and $\mathrm{R}$ are $1 \mathrm{~A}$ 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 $\sqrt{\mathrm{x} m T}$, where $\mathrm{x}=$
[Use $\mu_{0}=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}$ ]
Show Answer
Answer: (20)
Solution:
$\mathrm{B}{\mathrm{P}}=\frac{\mu{0} \mathrm{Ni}{1}}{2 \mathrm{r}}=\frac{\mu{0} \times 1 \times 100}{2 \pi}=2 \times 10^{-3} \mathrm{~T}$
$\mathrm{B}{\mathrm{Q}}=\frac{\mu{0} \mathrm{Ni}{2}}{2 \mathrm{r}}=\frac{\mu{0} \times 2 \times 100}{2 \pi}=4 \times 10^{-3} \mathrm{~T}$
$\mathrm{B}{\text {net }}=\sqrt{\mathrm{B}{\mathrm{P}}^{2}+\mathrm{B}_{\mathrm{Q}}^{2}}$
$=\sqrt{20} \mathrm{mT}$
$\mathrm{x}=20$
