Question

The wire loop PQRSP formed by joining two semicircular wires of radii $\sf  R_1$ and $\sf R_2$ carries a current $\sf I$ as shown. the magnitude of the magnetic induction at the centre C :

$\sf (A) \quad\dfrac{\mu_{0}I}{2}\bigg( \dfrac{R_2 - R_1}{R_2R_1}\bigg)$

$\sf (B) \quad \dfrac{\mu_{0}I}{2\pi}\bigg( \dfrac{R_2 - R_1}{R_2R_1}\bigg)$

$\sf (C) \quad \dfrac{\mu_{0}I}{4}\bigg( \dfrac{1}{R_2} - \dfrac{1}{R_1}\bigg)$

$\sf (D) \quad \dfrac{\mu_{0}I}{4\pi}\bigg(\dfrac{1}{R_2} - \dfrac{1}{R_1}\bigg)$

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Answer 1

Solution:

First of all, wires PQ and SR will not contribute any field intensity at point O, because their axes $\rm pa ss$ through point O.

So, field intensity will only be given by the semi-circular part.

$\sf B = \dfrac{ \mu_{0}}{4\pi} \bigg(\dfrac{\pi I}{r_{1}} \bigg) - \dfrac{ \mu_{0}}{4\pi} \bigg(\dfrac{\pi I}{r_{2}} \bigg)$

$\sf \implies B = \dfrac{ \mu_{0}}{4} \bigg(\dfrac{ I}{r_{1}} \bigg) - \dfrac{ \mu_{0}}{4} \bigg(\dfrac{I}{r_{2}} \bigg)$

$\sf \implies B = \dfrac{ \mu_{0}I}{4} \bigg(\dfrac{ I}{r_{1}} - \dfrac{1}{ r_{2}} \bigg)$

$\sf \implies B = \dfrac{ \mu_{0}I}{4} \bigg(\dfrac{ r_{2} - r_{1} }{ r_{1} r_{2}} \bigg)$

Hope It Helps.