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  • In the figure shown A B C D E F A was a square loop of side l, but is folded in two equal parts so that half of it lies in x z plane and the other half lies in th

In the figure shown A B C D E F A was a square loop of side l, but is folded in two equal parts so that half of it lies in x z plane and the other half lies in the y z plane. The origin ' O ' is centre of the frame also. The loop carries current ' i '. The magnetic field at the centre is: 

Option: 1

\frac{\mu_{0}i}{2\sqrt{2}\pi \l }\left ( \hat{i}-\hat{j} \right )


Option: 2

\frac{\mu_{0}i}{4\pi \l }\left ( -\hat{i}+\hat{j} \right )


Option: 3

\frac{\sqrt{2}\mu_{0}i}{\pi \l }\left ( \hat{i}+\hat{j} \right )


Option: 4

\frac{\mu_{0}i}{\sqrt{2}\pi \l }\left ( \hat{i}+\hat{j} \right )


Answers (1)

best_answer

Due to FABC the magnetic field at O is along y – axis and due to CDEF the magnetic field is along x–axisHence the field will be of the form A  \left [\vec{i}+\vec{j} \right ]

Calculating field due to FABC:

 due to AB :\vec{B}_{AB}=\frac{\mu _{0}i}{4\pi\left ( \frac{l}{2} \right )}\left ( sin 45^{\circ} +sin 45^{\circ} \right )\hat{j}=\sqrt{2}\frac{\mu _{0}i}{2\pi l}\hat{j}

 

Due to BC:\vec{B}_{Bc}=\frac{\mu _{0}i}{4\pi \left ( \frac{l}{2} \right )}\left ( sin 0^{\circ} +sin 45^{\circ} \right )\hat{j}=\frac{\mu _{0}i}{2\sqrt{2}\pi l}\hat{j}

Similarly due to FA : \vec{B}_{FA}=\frac{\mu _{0}i}{2\sqrt{2}\pi l}\hat{j}

Hence \vec{B}_{FABC}=\frac{\mu_{0}i}{\pi l}\left [ \frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}+\frac{\sqrt{2}}{2} \right ]\hat{i}=\vec{B}_{FABC}=\frac{\sqrt{2} \mu_{0}i}{\pi l}\left ( \vec{j} \right )

Similarly due to CDEF:

\vec{B}_{CDEF}=\frac{\sqrt{2} \mu_{0}i}{\pi l}\left ( \vec{i} \right )\Rightarrow \vec{B}_{net}=\frac{\sqrt{2} \mu_{0}i}{\pi l}\left ( \vec{i}+\vec{j} \right )

 

Posted by

Irshad Anwar

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