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)

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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