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A current of $1 \mathrm{~A}$ is flowing on the sides of an equilateral triangle of side $4.5 \times 10^{-2} \mathrm{~m}$. The magnetic field at the centre of the triangle will be:
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The correct answer is:
$4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$
$4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$
Here, side of the triangle, $l=4.5 \times 10^{-2} \mathrm{~m}$, current, $\mathrm{I}=1 \mathrm{~A}$
magnetic field at the centre of the triangle ' $O$ ' $B=$ ?
From figure, $\tan 60^{\circ}=\sqrt{3}=\frac{1}{2 d}$
$$
\Rightarrow d=\frac{l}{2 \sqrt{3}}=\left(\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}\right) \mathrm{m}
$$
Magnetic field, $B=\frac{\mu_0 i}{4 \pi d}\left(\cos \theta_1+\cos \theta_2\right)$
Putting value of $\mu=4 \pi \times 10^{-7}$ and $\theta_1$ and $\theta_2$ we will get net magenetic field $=3 \times B=4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$
magnetic field at the centre of the triangle ' $O$ ' $B=$ ?
From figure, $\tan 60^{\circ}=\sqrt{3}=\frac{1}{2 d}$
$$
\Rightarrow d=\frac{l}{2 \sqrt{3}}=\left(\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}\right) \mathrm{m}
$$
Magnetic field, $B=\frac{\mu_0 i}{4 \pi d}\left(\cos \theta_1+\cos \theta_2\right)$
Putting value of $\mu=4 \pi \times 10^{-7}$ and $\theta_1$ and $\theta_2$ we will get net magenetic field $=3 \times B=4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$
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