NEET 2023 — Physics solution

NEET2023Physics

A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by:

A
$\frac{μ _{0} i}{4 R}$ pointed away from the page
B
$\frac{μ _{0} i}{4 R} [1 - \frac{2}{π}]$ pointed away from the page
C
$\frac{μ _{0} i}{4 R} [1 - \frac{2}{π}]$ pointed into the page
D
$\frac{μ _{0} i}{4 R}$ pointed into the page

Solution

Field at the centre of a semicircular arc: $B_{a r c} = \frac{μ _{0} i}{4 R}$ (outward, by the right-hand rule for this geometry).

Each semi-infinite straight wire produces a field at the centre P (which lies on its extended line on one side) of magnitude $\frac{μ _{0} i}{4 π R}$ , directed into the page. Two such segments give a total of $\frac{μ _{0} i}{2 π R}$ into the page.

Net (taking outward positive): $B = \frac{μ _{0} i}{4 R} - \frac{μ _{0} i}{2 π R} = \frac{μ _{0} i}{4 R} [1 - \frac{2}{π}]$ , pointed away from the page.

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