JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
The current passing through a conducting loop in the form of equilateral triangle of side \(4\sqrt{3}\) cm is 2A. The magnetic field at its centroid is \(\alpha\times10^{-5}T\). The value of \(\alpha\) is ___________.
(Given : \(\mu_{o}=4\pi\times10^{-7}\) SI units)
- A \(2 \sqrt{3}\)
- B \(\sqrt{3}\)
- C \(3 \sqrt{3}\)
- D \(\frac{\sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(C) \(3 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(B =\frac{\mu_0}{4 \pi} \times \frac{ I }{ d }\left[\sin 60^{\circ}+\sin 60^{\circ}\right] \times 3\) \(B =10^{-7} \times \frac{2}{2 \times 10^{-2}}\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right) \times 3\) \(=\sqrt{3} \times 10^{-5} \times 3=3 \sqrt{3} \times 10^{-5}\)
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