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  • Help me please, - Magnetic Effects of Current and Magnetism - BITSAT

Two identical wires A and B, each of length ‘l’, carry the same current I.  Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’.  If BA and BB are the values of magnetic field at the centres of the circle and square
respectively, then the ratio\frac{B_{A}}{B_{B}} is

  • Option 1)

    \frac{\pi ^{2}}{ }

  • Option 2)

    \frac{\pi ^{2}}{16\sqrt{2}}

  • Option 3)

    \frac{\pi ^{2}}{16}

  • Option 4)

    \frac{\pi ^{2}}{8\sqrt{2}}

 

Answers (1)

As we have learnt,

 

Magnetic Field due to circular coil at Centre -

B_{Centre}=\frac{\mu_{0}}{4\pi }\frac{2\pi Ni}{r}=\frac{\mu_{0} Ni}{2r}

- wherein

X = 0

 

 Magnetic field at the centre of circle 

B = \frac{\mu_o I}{2r} \;\;\;\because 2\pi r = l

                         \therefore 2r = \frac{l}{\pi}

B = \frac{\mu_o I \pi}{l}

in case of square

B =4\times \frac{\mu_o I}{4\pi\cdot \frac{a}{2}}(\cos45\degree + \sin45\degree)

\\ =4\times \frac{\mu_o I}{2\pi{a}}\cdot \frac{2}{\sqrt2} \\ \therefore 4 a = l \Rightarrow a = \frac{l}{4}

B = 4\cdot \frac{\mu_o I}{2\pi\left(\frac{l}{4} \right )}\cdot \sqrt2 = 8\sqrt2\cdot\frac{\mu_o I}{2\pi l}

\frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt2}


Option 1)

\frac{\pi ^{2}}{ }

Option 2)

\frac{\pi ^{2}}{16\sqrt{2}}

Option 3)

\frac{\pi ^{2}}{16}

Option 4)

\frac{\pi ^{2}}{8\sqrt{2}}

Posted by

Vakul

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