JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 2)
A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is :
(Assume that the current is flowing in the clockwise direction.)
(Assume that the current is flowing in the clockwise direction.)
3 $$\times$$ 10$$-$$7 T, outside the plane of triangle
$$2\sqrt 3 $$ $$\times$$ 10$$-$$7 T, outside the plane of triangle
$$2\sqrt 3 $$ $$\times$$ 10$$-$$5 T, inside the plane of triangle
3 $$\times$$ 10$$-$$5 T, inside the plane of triangle
Explanation
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$$B = 3\left[ {{{{\mu _0}i} \over {4\pi r}}(\sin 60^\circ + \sin 60^\circ )} \right]$$
$$\tan 60^\circ = {{l/2} \over r}$$
where $$r = {{9 \times {{10}^{ - 2}}} \over {2\sqrt 3 }}$$ M
$$\therefore$$ B = 3 $$\times$$ 10-5 T
Current is flowing in clockwise direction so, $$\overrightarrow B $$ is inside plane of triangle by right hand rule.
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