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

To resolve the problem, we commence by computing the magnetic field at the center of each coil, generated by the currents therein. The magnetic field \( B \) at the center of a single circular coil, characterized by \( N \) turns, radius \( r \), and current \( I \), is determined by the formula \( B = \frac{\mu_0 N I}{2r} \). Both coils share a common radius \( r = \pi \, \text{cm} = 0.01\pi \, \text{m} \).
For coil \( P \), with current \( I_P = 1 \, \text{A} \):
\( B_P = \frac{(4\pi \times 10^{-7}) \times 100 \times 1}{2 \times 0.01\pi} \). Upon simplification, \( B_P = 2 \times 10^{-3} \, \text{T} \), which is equivalent to \( 2 \, \text{mT} \).
For coil \( Q \), with current \( I_Q = 2 \, \text{A} \): 
\( B_Q = \frac{(4\pi \times 10^{-7}) \times 100 \times 2}{2 \times 0.01\pi} \). After simplification, \( B_Q = 4 \times 10^{-3} \, \text{T} \), or \( 4 \, \text{mT} \).
The magnetic fields \( B_P \) and \( B_Q \) are oriented perpendicularly to each other. The resultant magnetic field \( B_R \) at the center is calculated using the Pythagorean theorem: \( B_R = \sqrt{B_P^2 + B_Q^2} = \sqrt{2^2 + 4^2} = \sqrt{20} \, \text{mT} \).
Consequently, the value of \( x \) is \( 20 \). This value is within the stipulated range of 20,20.