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 $4 a$ . The magnitude of the magnetic field at the center of the loop is [2017]

Question

Smart Achievers · Accepted Answer

(1)

From figure,

In $∆ O A C$ ,

$\cos 60 ° = \frac{O C}{O A} = \frac{O C}{2 a}$

$∴$ $O C = 2 a \times \frac{1}{2} = a$

Magnetic field at $' O'$ due to element AB

$= \frac{μ_{0}}{4 π} \frac{I}{a} (\sin 60 ° - \sin 30 °)$

$= \frac{μ_{0}}{4 π} \frac{I}{a} (\frac{\sqrt{3}}{2} - \frac{1}{2})$ $= \frac{μ_{0} I}{4 π a} \times \frac{1}{2} (\sqrt{3} - 1)$

$∴$ Magnetic field at the centre, due to the complete loop

$= [\frac{μ_{0} I}{4 π a} \times \frac{1}{2} (\sqrt{3} - 1)] \times 12$ $= \frac{μ_{0} I}{4 π a} 6 (\sqrt{3} - 1)$

Solution

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 $4 a$ . The magnitude of the magnetic field at the center of the loop is [2017]

1 $\frac{μ_{0} I}{4 π a} 6 [\sqrt{3} - 1]$

2 $\frac{μ_{0} I}{4 π a} 6 [\sqrt{3} + 1]$

3 $\frac{μ_{0} I}{4 π a} 3 [\sqrt{3} - 1]$

4 $\frac{μ_{0} I}{4 π a} 3 [2 - \sqrt{3}]$

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Solution

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. The magnitude of the magnetic field at the center of the loop is [2017]

1 μ0I4πa6[3-1]

2 μ0I4πa6[3+1]

3 μ0I4πa3[3-1]

4 μ0I4πa3[2-3]