A very long wire ABDMNDC is shown in figure carrying current I. AB… | JEE Main 2020 (8 Jan, Shift 2)

Q55JEE Main 2020 · 8 JAN · SHIFT 2Medium

Magnetic Field of Straight Wires

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, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is:

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Answer: (d)

AB (from $-\infty$ to B, foot N is $R$ beyond B): $\frac{\mu_0 I}{4\pi R}\left(1 - \frac{1}{\sqrt{2}}\right)$ , into the page.
BC (from B to $\infty$ , foot D is $R$ beyond B): $\frac{\mu_0 I}{4\pi R}\left(1 + \frac{1}{\sqrt{2}}\right)$ , out of the page.
Wires net: $\frac{\mu_0 I}{4\pi R}\sqrt{2} = \frac{\mu_0 I}{2\pi R}\cdot\frac{1}{\sqrt{2}}$ , out of the page.
Circular turn (anticlockwise): $\frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2\pi R}\pi$ , out of the page.
$B = \frac{\mu_0 I}{2\pi R}\left(\pi + \frac{1}{\sqrt{2}}\right)$ , so the answer is (d).

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Magnetic Field of Straight Wires question from JEE Main 2020 (8 Jan, Shift 2)

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