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 \(\mu_0 = 4 \pi \times 10^{-7} \, \text{TmA}^{-1}\)]
[Use \(\mu_0 = 4 \pi \times 10^{-7} \, \text{TmA}^{-1}\)]
Correct Answer: 20
Approach Solution - 1
To solve the problem, we first calculate the magnetic field at the center of each coil due to the currents flowing through them. The magnetic field \( B \) at the center of a single circular coil with \( N \) turns, radius \( r \), and current \( I \) is given by the formula: \( B = \frac{\mu_0 N I}{2r} \). Both coils have the same radius \( r = \pi \, \text{cm} = 0.01\pi \, \text{m} \).
For coil \( P \) (with current \( I_P = 1 \, \text{A} \)):
\( B_P = \frac{(4\pi \times 10^{-7}) \times 100 \times 1}{2 \times 0.01\pi} \). Simplifying, \( B_P = 2 \times 10^{-3} \, \text{T} = 2 \, \text{mT} \).
For coil \( Q \) (with current \( I_Q = 2 \, \text{A} \)):
\( B_Q = \frac{(4\pi \times 10^{-7}) \times 100 \times 2}{2 \times 0.01\pi} \). Simplifying, \( B_Q = 4 \times 10^{-3} \, \text{T} = 4 \, \text{mT} \).
The magnetic fields \( B_P \) and \( B_Q \) are perpendicular to each other. The resultant magnetic field \( B_R \) at the center is found using the Pythagorean theorem: \( B_R = \sqrt{B_P^2 + B_Q^2} = \sqrt{2^2 + 4^2} = \sqrt{20} \, \text{mT} \).
Thus, the value of \( x \) is \( 20 \). This value falls within the expected range of 20,20.
Approach Solution -2
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, \( x = 20 \).
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