A regular polygon of 6 sides is formed by bending a wire of length 4 π meter. If an electric current of 4 π √(3) A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be x × 10 -7 T . The value of x is \ \ \ \ \ \ \ ・

Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 1 February, Shift 1 — Question 47

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 $\mathrm{x} \times 10^{-7} \mathrm{~T}$ . The value of x is $\_\_\_\_\_\_\_$ ・

Answer: 72

Numerical answer — enter this value.

Step-by-step solution

$\begin{array}{*{35}{r}}{} & B=6\left( \frac{{{\mu }_{0}}I}{4\pi r} \right)\left( \text{sin}{{30}^{\circ }}+\text{sin}{{30}^{\circ }} \right) \\{} & ~=6\frac{{{10}^{-7}}\times 4\pi \sqrt{3}}{\left( \frac{\sqrt{3}\times 4\pi }{2\times 6} \right)} \\{} & ~=72\times {{10}^{-7}}\text{ }\!\!~\!\!\text{ T} \\\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law