A regular polygon of 6 sides is formed by bending a wire of length 4 \pi meter. If an electric current of 4 \pi \sqrt3 A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be x \times 10^-7 ~T . The value of x is \underline\hspace2cm .

Magnetics PYQ — JEE Main 2024 · 1 Feb · Shift 1 · Q86 with Solution

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Correct answer: 72

Identify the polygon and side length

A regular polygon of $6$ sides is a regular hexagon.

Total wire length = perimeter = $4\pi$ m.

So, side length of hexagon is

a = \frac{4\pi}{6} = \frac{2\pi}{3} \text{ m}

Magnetic field at the centre due to one side

For a finite straight wire, magnetic field at a point at perpendicular distance $r$ is

B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)

At the centre of a regular hexagon, for one side:

the perpendicular distance from centre to side is the apothem,
the wire subtends equal angles at the centre, so $\theta_1 = \theta_2 = \theta$ .

For a regular $n$ -gon,

\theta = \frac{\pi}{n}

So for $n=6$ ,

\theta = \frac{\pi}{6}

Thus field due to one side is

B_1 = \frac{\mu_0 I}{4\pi r}(2\sin\tfrac{\pi}{6})

Since $\sin \frac{\pi}{6} = \frac12$ ,

B_1 = \frac{\mu_0 I}{4\pi r}

Find the apothem $r$

For a regular hexagon, circumradius equals side length:

R=a=\frac{2\pi}{3}

Apothem is

r = R\cos\frac{\pi}{6} = a\cdot \frac{\sqrt{3}}{2}

r = \frac{2\pi}{3}\cdot \frac{\sqrt{3}}{2} = \frac{\pi\sqrt{3}}{3}

Field due to all 6 sides

All contributions are in the same direction, so

B = 6B_1 = 6\cdot \frac{\mu_0 I}{4\pi r}

Given current

I = 4\pi\sqrt{3}\ \text{A}

and

\mu_0 = 4\pi\times 10^{-7}

Hence

B = 6\cdot \frac{4\pi\times 10^{-7}\cdot 4\pi\sqrt{3}}{4\pi\cdot (\pi\sqrt{3}/3)}

Simplify:

B = 6\cdot 10^{-7}\cdot \frac{4\pi\sqrt{3}}{\pi\sqrt{3}/3}

B = 6\cdot 10^{-7}\cdot 12

B = 72\times 10^{-7}\ \text{T}

Final value

Comparing with

B = x\times 10^{-7}\ \text{T}

we get

x = 72

Comparison with stored answer

Stored correct answer = $72$ .

This matches our derived answer.

More from Magnetics

View written solutionFree

Correct answer: 72

Identify the polygon and side length

A regular polygon of $6$ sides is a regular hexagon.

Total wire length = perimeter = $4\pi$ m.

So, side length of hexagon is

a = \frac{4\pi}{6} = \frac{2\pi}{3} \text{ m}

Magnetic field at the centre due to one side

For a finite straight wire, magnetic field at a point at perpendicular distance $r$ is

B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)

At the centre of a regular hexagon, for one side:

the perpendicular distance from centre to side is the apothem,
the wire subtends equal angles at the centre, so $\theta_1 = \theta_2 = \theta$ .

For a regular $n$ -gon,

\theta = \frac{\pi}{n}

So for $n=6$ ,

\theta = \frac{\pi}{6}

Thus field due to one side is

B_1 = \frac{\mu_0 I}{4\pi r}(2\sin\tfrac{\pi}{6})

Since $\sin \frac{\pi}{6} = \frac12$ ,

B_1 = \frac{\mu_0 I}{4\pi r}

Find the apothem $r$

For a regular hexagon, circumradius equals side length:

R=a=\frac{2\pi}{3}

Apothem is

r = R\cos\frac{\pi}{6} = a\cdot \frac{\sqrt{3}}{2}

r = \frac{2\pi}{3}\cdot \frac{\sqrt{3}}{2} = \frac{\pi\sqrt{3}}{3}

Field due to all 6 sides

All contributions are in the same direction, so

B = 6B_1 = 6\cdot \frac{\mu_0 I}{4\pi r}

Given current

I = 4\pi\sqrt{3}\ \text{A}

and

\mu_0 = 4\pi\times 10^{-7}

Hence

B = 6\cdot \frac{4\pi\times 10^{-7}\cdot 4\pi\sqrt{3}}{4\pi\cdot (\pi\sqrt{3}/3)}

Simplify:

B = 6\cdot 10^{-7}\cdot \frac{4\pi\sqrt{3}}{\pi\sqrt{3}/3}

B = 6\cdot 10^{-7}\cdot 12

B = 72\times 10^{-7}\ \text{T}

Final value

Comparing with

B = x\times 10^{-7}\ \text{T}

we get

x = 72

Comparison with stored answer

Stored correct answer = $72$ .

This matches our derived answer.

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