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JEE Advanced 2017 · Paper 2, Physics question 4

A symmetric star shaped conducting wire loop is carrying a steady state current II as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a4a. The magnitude of the magnetic field at the center of the loop is

A six-pointed star (hexagram outline) wire loop carrying current I; the distance between opposite outer vertices (top to bottom) is 4a.
Choose one option
Answer and worked solution

Official answer: A

Geometry. The hexagram is made of two equilateral triangles with circumradius 2a2a. Each triangle side lies at distance aa from the centre and has half-length 3a\sqrt3a. The loop's 12 edges each run, along such a side, from an inner vertex (at a/3a/\sqrt3 from the foot of the perpendicular) to an outer vertex (at 3a\sqrt3a).

One segment. Using B=μ0I4πd(sin⁡θ2−sin⁡θ1)B=\dfrac{\mu_0I}{4\pi d}(\sin\theta_2-\sin\theta_1), angles measured from the perpendicular:
sin⁡θ2=3a2a=32,sin⁡θ1=a/32a/3=12.\sin\theta_2=\frac{\sqrt3a}{2a}=\frac{\sqrt3}2,\qquad\sin\theta_1=\frac{a/\sqrt3}{2a/\sqrt3}=\frac12.
So
Bseg=μ0I4πa⋅3−12.B_{seg}=\frac{\mu_0I}{4\pi a}\cdot\frac{\sqrt3-1}2.

All 12 segments add in the same direction:
B=μ0I4πa 6(3−1).B=\frac{\mu_0I}{4\pi a}\,6(\sqrt3-1).

(A)

Topics:Biot-Savart law
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