Question

Two circular coils $P$ and $Q$ of $100$ turns each have the same radius of $\pi \, \text{cm}$. The currents in $P$ and $Q$ are $1 \, \text{A}$ and $2 \, \text{A}$ respectively. $P$ and $Q$ are placed with their planes mutually perpendicular with their centers coinciding. The resultant magnetic field induction at the center of the coils is $\sqrt{x} \, \text{mT}$, where $x =$ ______.
[Use $\mu_0 = 4 \pi \times 10^{-7} \, \text{TmA}^{-1}$]

Answer 1

Number of turns: $N = 100$
Radius of coils: $r = \pi \, \text{cm} = \pi \times 10^{-2} \, \text{m}$
Current in coil $P$: $I_1 = 1 \, \text{A}$
Current in coil $Q$: $I_2 = 2 \, \text{A}$

The magnetic field at the center of a circular coil is given by:

$B = \frac{\mu_0 NI}{2r}$ where $\mu_0 = 4\pi \times 10^{-7} \, \text{TmA}^{-1}$.

Calculating the magnetic fields:

$B_P = \frac{\mu_0 NI_1}{2r} = \frac{4\pi \times 10^{-7} \times 100 \times 1}{2 \times \pi \times 10^{-2}} = 2 \times 10^{-3} \, \text{T}$ $B_Q = \frac{\mu_0 NI_2}{2r} = \frac{4\pi \times 10^{-7} \times 100 \times 2}{2 \times \pi \times 10^{-2}} = 4 \times 10^{-3} \, \text{T}$

Since the magnetic fields are perpendicular, the resultant magnetic field $B_{\text{net}}$ is given by:

$B_{\text{net}} = \sqrt{B_P^2 + B_Q^2}$ $B_{\text{net}} = \sqrt{(2 \times 10^{-3})^2 + (4 \times 10^{-3})^2} \, \text{T}$ $B_{\text{net}} = \sqrt{4 \times 10^{-6} + 16 \times 10^{-6}} \, \text{T}$ $B_{\text{net}} = \sqrt{20 \times 10^{-6}} \, \text{T}$ $B_{\text{net}} = \sqrt{20} \times 10^{-3} \, \text{T} = \sqrt{20} \, \text{mT}$
Thus, $x = 20$.