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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2020
  • Hard
  • MCQ

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

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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 \((A, B, C)\)

\(\mathrm{B}=\mathrm{B}_{\mathrm{A}}+\mathrm{B}_{\mathrm{B}}+\mathrm{B}_{\mathrm{C}}\)

\(=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\sin 90^{\circ}-\sin 45^{\circ}\right) \otimes+\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}} \odot+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\sin 45^{\circ}+\sin 90^{\circ}\right) \odot\)

\(=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{R}}\left(\sin 45^{\circ}+\pi\right)\)

\(=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{R}}\left(\pi+\frac{1}{\sqrt{2}}\right)\)

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