JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2024
- Hard
- Numerical
A regular polygon of \(6\) sides is formed by bending a wire of length \(4 \pi\) meter. If an electric current of \(4 \pi \sqrt{3} \mathrm{~A}\) is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be \(x \times 10^{7} \mathrm{~T}\). The value of \(\mathrm{x}\) is ______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(72\)
Step-by-step Solution
Detailed explanation
\(s = \frac{P}{n} = \frac{4\pi}{6} = \frac{2\pi}{3} \mathrm{~m}\)
\(a = \frac{s}{2 \tan(\pi/n)} = \frac{2\pi/3}{2 \tan(\pi/6)} = \frac{2\pi/3}{2(1/\sqrt{3})} = \frac{\pi\sqrt{3}}{3} \mathrm{~m}\)
\(B = \frac{n \mu_0 I}{2 \pi a} \sin(\pi/n) = \frac{6 \times (4\pi \times 10^{-7}) \times (4\pi\sqrt{3})}{2 \pi \times (\pi\sqrt{3}/3)} \times \sin(\pi/6)\)
\(B = \frac{6 \times 4\pi \times 10^{-7} \times 4\pi\sqrt{3}}{2 \pi^2 \sqrt{3}/3} \times \frac{1}{2}\)
\(B = \frac{96 \pi^2 \sqrt{3} \times 10^{-7}}{2 \pi^2 \sqrt{3}/3} \times \frac{1}{2}\)
\(B = 72 \times 10^{-7} \mathrm{~T}\)
\(x = 72\)

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