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

Correct Answer: 20

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, \( x = 20 \).