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A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 ×10^{-2} m. The magnetic field at the centre of the triangle will be :

  • Option 1)

    2 ×10^{-5} Wb/m^{2}

  • Option 2)

    Zero

  • Option 3)

    8 ×10^{-5} Wb/m^{2}

  • Option 4)

    4 ×10^{-5} Wb/m^{2}

 

Answers (1)

As we learnt

Magnetic Field Due to a Straight Wire -

B=\frac{\mu_{o}i}{4\pi r}(\sin\phi_{1}+\sin\phi_{2})

- wherein

 

                                

 

 

 \overrightarrow{B}_{BC}=\frac{\mu _{0}I}{4\Pi\frac{a}{2\sqrt{3}} } \left ( sin60^{\circ}+sin60^{\circ} \right )

              =\frac{\mu _{0}I}{2\Pi{a}}\sqrt{3}\left ( \sqrt{3} \right )=\frac{3\mu _{0}I}{2\Pi{a}}

               =\frac{3*2*10^{-7*1}}{4.5*10^{-2}}=\frac{2}{1.5}*10^{-5}Wb/m^{2}

\thereforeTotal Magnetic field   =\frac{3*2*10^{-5}}{1.5}

                                      =4*10^{-5}Wb/m^{2}

 


Option 1)

2 ×10^{-5} Wb/m^{2}

2

Option 2)

Zero

Option 3)

8 ×10^{-5} Wb/m^{2}

Option 4)

4 ×10^{-5} Wb/m^{2}

Posted by

Vakul

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