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JEE Advance - Physics (2017 - Paper 2 Offline - No. 9)

A symmetric star shaped conducting wire loop is carrying a steady state current $${\rm I}$$ 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

JEE Advanced 2017 Paper 2 Offline Physics - Magnetism Question 40 English
$${{{\mu _0}1} \over {4\pi a}}6\left[ {\sqrt 3 - 1} \right]$$
$${{{\mu _0}1} \over {4\pi a}}6\left[ {\sqrt 3 + 1} \right]$$
$${{{\mu _0}1} \over {4\pi a}}3\left[ {\sqrt 3 - 1} \right]$$
$${{{\mu _0}1} \over {4\pi a}}3\left[ {2 - \sqrt 3 } \right]$$

Explanation

The star shape is composed of 12 wires. Thus, the total magnetic field at centre is 12 times of magnetic field due to one wire.

JEE Advanced 2017 Paper 2 Offline Physics - Magnetism Question 40 English Explanation

Let us first calculate the magnetic field due to element AB. Perpendicular distance of centre O from element AB, OP = a

Angle subtended by centre at element are $$\theta$$1 = 30$$^\circ$$, $$\theta$$2 = 60$$^\circ$$ as shown in figure.

$$\therefore$$ $$\overrightarrow B = {{{\mu _0}I} \over {4\pi a}}(\cos 30^\circ - \cos 60^\circ ) \odot $$

$$ = {{{\mu _0}I} \over {4\pi a}}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right) \odot = {{{\mu _0}I} \over {8\pi a}}(\sqrt 3 - 1) \odot $$

Since the direction of magnetic field due to each element side is out of the paper

$$\therefore$$ Net magnetic field at O $$ = 12 \times {{{\mu _0}I} \over {8\pi a}}(\sqrt 3 - 1)$$

$$ = {{{\mu _0}I} \over {4\pi a}}6(\sqrt 3 - 1)$$

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