A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a .

Question

A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-r8mnfhphg2gi.png" /> (A) (μ 0 I/4 π a) 3 [√3 - 1] (B) (μ 0 I/4 π a)6[√3 - 1] (C) (μ 0 I/4 π a)6 [√3 + 1] (D) (μ 0 I/4 π a)3[2 - √3]

Tardigrade Admin · Accepted Answer

The given points (1, 2, 3, 4, 5, 6) makes 360^o angle at ‘O’. Hence angle made by vertices 1 and 2 with ‘O’ is 60^o.
Solution

Direction of magnetic field at ‘O’ due to each segment is same. Since it is symmetric star shape, magnitude will also be same.
Magnetic field due to section BC.

(B_{1}) = \frac{ki}{a} (sin (6 0^{\circ}) - sin 3 0^{\circ}) = \frac{ki}{2 a} (3 - 1)

B_{n e t} = 12 \times B_{1} = \frac{6 ki}{a} (3 - 1)

and

(k = \frac{μ _{0}}{4 π})

Q. A symmetric star shaped conducting wire loop is carrying a steady state current $I$ as shown in the figure. The distance between the diametrically opposite vertices of the star is $4 a$ .

$\frac{μ _{0} I}{4 πa} 3 [3 - 1]$

$\frac{μ _{0} I}{4 πa} 6 [3 - 1]$

$\frac{μ _{0} I}{4 πa} 6 [3 + 1]$

$\frac{μ _{0} I}{4 πa} 3 [2 - 3]$

Q. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a .

4πaμ0​I​3[3​−1]

4πaμ0​I​6[3​−1]

4πaμ0​I​6[3​+1]

4πaμ0​I​3[2−3​]

Q. A symmetric star shaped conducting wire loop is carrying a steady state current $I$ as shown in the figure. The distance between the diametrically opposite vertices of the star is $4 a$ .

$\frac{μ _{0} I}{4 πa} 3 [3 - 1]$

$\frac{μ _{0} I}{4 πa} 6 [3 - 1]$

$\frac{μ _{0} I}{4 πa} 6 [3 + 1]$

$\frac{μ _{0} I}{4 πa} 3 [2 - 3]$