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  • The magnitude of magnetic induction at mid-point O due to the current arrangement as shown in Fig will be  <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/05/19/captu

The magnitude of magnetic induction at mid-point O due to the current arrangement as shown in Fig will be

 

Option: 1

\frac{\mu_oI}{\pi a}


Option: 2

\frac{\mu_oI}{2\pi a}


Option: 3

0


Option: 4

\frac{\mu_oI}{4\pi a}


Answers (1)

best_answer

Magnetic field due to "AB" and "ED" will be zero
magnetic field due to "BC" and "ET" will be equal in amount and direction.

'B' \ due \ \mathrm{BC}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{r}}=\frac{\mu_0 \mathrm{I}}{4 \pi \frac{\mathrm{a}}{2}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}}\\ ' \mathrm{B} ' \ due \ to \ \mathrm{TE}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \odot\\ \mathrm{B}_{\text {net }} \ at \ point ' \ \mathrm{O}^{\prime}=\left(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}}+\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}}\right)=\frac{\mu_0 \mathrm{I}}{\pi \mathrm{a}} \odot outward

Posted by

sudhir.kumar

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