JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A very long wire \(ABDMNDC\) is shown in figure carrying current \(I. AB\) and \(BC\) parts are straight, long and at right angle. At \(D\) wire forms a circular turn \(DMND\) of radius \(R. AB.\) \(\mathrm{BC}\) parts are tangential to circular turn at \(\mathrm{N}\) and \(D\). Magnetic field at the centre of circle is

- A \(\frac{\mu_{0} I}{2 R}\)
- B \(\frac{\mu_{0} I}{2 \pi R}(\pi+1)\)
- C \(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{R}}\left(\pi+\frac{1}{\sqrt{2}}\right)\)
- D \(\frac{\mu_{0} I}{2 \pi R}\left(\pi-\frac{1}{\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{R}}\left(\pi+\frac{1}{\sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
We say we have \(3\) parts \((A, B, C)\) \(\mathrm{B}=\mathrm{B}_{\mathrm{A}}+\mathrm{B}_{\mathrm{B}}+\mathrm{B}_{\mathrm{C}}\)…
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