QJEEM18Ph8bd5: JEE Main 2018 Physics - Class 12 JEEM18Ph8bd5 Solution: A current of 1 A is flowing on the sides of an equilateral…

We know that magnetic field due to finite current carrying wire is

$B = {{{\mu _0}} \over {4\pi }}{I \over b}(\cos {\theta _1} + \cos {\theta _2})$ ..... (1)

Given that side of triangle = 4.5 $\times$ 10 ^{$-$ 2} m = a; current = 1A since the triangle is equilateral, angle of each side will be 60 $^\circ$ .

Now, $\tan \theta = {{Perpendicular} \over {Base}} \Rightarrow \tan 60^\circ = {{a/2} \over b}$

$\Rightarrow b = {a \over {2\tan 60^\circ }} = {a \over {2\sqrt 3 }}$

Using equation (1), we get

$B = {{{\mu _0}} \over {4\pi }}{I \over {a/2\sqrt 3 }}(\cos 30^\circ + \cos 30^\circ ) = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}2\cos 30^\circ$

$B = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}{{2 \times \sqrt 3 } \over 2} = {{{\mu _0}} \over {4\pi }}{{6I} \over a}$

$\Rightarrow B = {{{{10}^{ - 7}} \times 6 \times 1} \over {4.5 \times {{10}^{ - 2}}}} = 1.33 \times {10^{ - 5}} \sim 2 \times {10^{ - 5}}$ Wb/m²