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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
$${{{\mu _0}1} \over {4\pi a}}6\left[ {\sqrt 3 - 1} \right]$$
$${{{\mu _0}1} \over {4\pi a}}6\left[ {\sqrt 3 + 1} \right]$$
$${{{\mu _0}1} \over {4\pi a}}3\left[ {\sqrt 3 - 1} \right]$$
$${{{\mu _0}1} \over {4\pi a}}3\left[ {2 - \sqrt 3 } \right]$$

Giải thích

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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