A current of 5A exists in a square loop of side 1 / √2 m . Then the magnitude of the magnetic field

Question

A current of 5 A exists in a square loop of side \(\frac{1}{\sqrt{2}} \mathrm{m}\). Then the magnitude of the magnetic field B at the centre of the square loop will be \(\mathrm{p} \times 10^{-6} \ \mathrm{T}\). where, value of p is _______.

[Take \(\mu_{0}=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}\)].

Answer 1

Answer is : 8

 

Let B be the magnetic field due to single side

then \(B=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}\left(\sin \theta_{1}+\sin \theta_{2}\right)\) 

\( =\frac{10^{-7} \times 5 \times 2}{\frac{1}{2 \sqrt{2}}} \times \frac{1}{\sqrt{2}}=2 \times 10^{-6}\) 

\(\therefore \ \mathrm{B}_{\text {net }}\) at centre \(\mathrm{O}=4 \mathrm{B}\) 

\( =8 \times 10^{-6}\) 

\( \therefore \ \mathrm{P}=8\)