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  • The magnetic field at the point of intersection of diagonals of square loop of side L carrying a current I is.Option: 1 <img alt="\mathrm{\

The magnetic field at the point of intersection of diagonals of square loop of side L carrying a current I is.

Option: 1

\mathrm{\frac{\mu_0 I}{\pi L} }


Option: 2

\mathrm{\frac{2 \mu_0 I}{\pi L}}


Option: 3

\mathrm{\frac{\sqrt{2} \mu_0 I}{\pi L}}


Option: 4

\mathrm{\frac{2 \sqrt{2} \mu_0 I}{\pi L}}


Answers (1)

best_answer

The magnetic field at the centre O due to the current element AE is given by
              \mathrm{ B_{A E}=-\frac{\mu_0 I}{4 \pi r} \int_{90^{\circ}}^{45^{\circ}} \sin \theta d \theta }

                        \mathrm{ =\frac{\mu_0 I}{4 \pi r}|\cos \theta|_{90^{\circ}}^{45^{\circ}} \\ }

                        \mathrm{ =\frac{\mu_0 I}{4 \pi r}\left(\cos 45^{\circ}-\cos 90^{\circ}\right) \\ }

                        \mathrm{ =\frac{\mu_0 I}{4 \pi r}\left(\cos 45^{\circ}-0\right)=\frac{\mu_0 I}{4 \sqrt{2} \pi r} }

It is clear that the magnetic field at O due to current element DE is the same as that due to AE.

Hence, the magnetic field at O due to one side AD is
              \mathrm{ B_{A D}=\frac{2 \mu_0 I}{4 \sqrt{2} \pi r}=\frac{\sqrt{2} \mu_0 I}{4 \pi r} }

Since the centre of the square is equidistant from the ends A,B,C and D of each side of the square and each side produces at the centre O the same magnetic field, the field due to the square is 4 times that due to one side. Hence (because r = L /2)
                    \mathrm{ B=4 B_{\mathrm{AD}}=\frac{\sqrt{2} \mu_0 I}{\pi r}=\frac{2 \sqrt{2} \mu_0 I}{\pi L} }

Posted by

Divya Prakash Singh

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