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The magnitude of the megnetic field at the center of an equilateral triangular loop of side 1 m Which is carrying a current of 10 A is;

$\left [ Take \mu _{0} =4\pi \times 10^{-7}NA^{-2}\right ]$

Option 1)
$18\mu T$
Option 2)
$9\mu T$
Option 3)
$3\mu T$
Option 4)
$1\mu T$

Answers (1)

best_answer

$a=1m$

$i=10A$

magnetic field by AB wire at O is

$B=\frac{\mu _{0}i}{4\pi l}\left [ \sin \theta _{1}+\sin \theta _{2} \right ]$

$B=\frac{\mu _{0}i}{4\pi l}\left [ \sin 60+\sin \ 60 \right ]$

$\frac{\mu _{0}}{4\pi }\times \frac{2\sqrt{3}\times 10}{l}\left [ \sqrt{3/2} +\sqrt{3/2}\right ]$

$B=\frac{\mu _{0}}{4\pi\, l }\times 30$

Net magnetic field at center =3B

$B_{net}=3\frac{\mu _{0}}{4\pi\, l }\times 30$

$=90 \times \frac{\mu _{0}}{2\pi\, \left ( 1 \right ) }$

$=90 \times \frac{4\pi \times 10^{-7}}{2\pi }$

$=18\times 10^{-6}$

$B_{net}=18\mu T$

Option 1)
$18\mu T$
Option 2)

$9\mu T$

Option 3)

$3\mu T$

Option 4)

$1\mu T$

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