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Question

A current is flowing in a hexagonal coil of side d. The magnetic induction at the center of the coil will be:

A
33μiπd
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B
3μi2πd
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C
μi33πd
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D
3μiπd
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Solution

The correct option is D √3μiπdLet us consider one side of regular hexagon QR as shown in the diagram. We want to calculate magnetic field at a point O, center of hexagon, such that OQR is an equilateral triangle.Taking a small element of length dl along QR at a distance of r from O then according to Biot-Savart law, we havedB=μidlsinθ4πr2Also, ∠R=600 and ∠PQO=1200 where P is the point in the line QR above Q.The resultant field B at point O is obtained by integrating the field due to all the elements from Q to R.Now, r=scosecθ where s is the distance between center O and mid point of QR.l=scotθ where l is the length of conductor from center of QR to element dlOr, dl=−s cosec2θdθFrom Biot Savart law, we havedB=−μiscosec2dθsinθ4πs2cosec2θ=−μisinθdθ4πsIntegrating above, we getB=∫6001200−μisinθdθ4πsB=μi(cos600−cos1200)4πs ........................ (1)Also, s=√3d2 ............................... (2)∴ from (1) and (2), B=μi2√3πd∴ for 6 sides, we haveB′=6μi2√3πd=√3μiπd

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