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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A current of \(1\,A\) is flowing on the sides of an equilateral triangle of side \(4.5\times10^{-2}\,m\) . The magnetic field at the centre of the triangle will be
- A \(4\times10^{-5}\,Wb/m^2\)
- B Zero
- C \(2\times10^{-5}\,Wb/m^2\)
- D \(8\times10^{-5}\,Wb/m^2\)
Answer & Solution
Correct Answer
(A) \(4\times10^{-5}\,Wb/m^2\)
Step-by-step Solution
Detailed explanation
Here, side of the triangle, \(l=4.5 \times 10^{-2} \,\mathrm{m}\) current, \(I =1 \,\mathrm{A}\) magnetic field at the centre of the triangle \('O'B = ?\) From figure, \(\tan 60^{\circ}=\sqrt{3}=\frac{1}{2 d}\)…
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