Figure shows a current carrying square loop ABCD of edge length is 'a' lying in a plane. If the resistance of the ABC part is r and that of ADC part is 2r, then the magnitude of the resultant magnetic field at centre of the square loop is [2025]

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Solution

Q.
Figure shows a current carrying square loop ABCD of edge length is 'a' lying in a plane. If the resistance of the ABC part is r and that of ADC part is 2r, then the magnitude of the resultant magnetic field at centre of the square loop is [2025]

1 $\frac{3 π μ_{0} I}{\sqrt{2} a}$

2 $\frac{μ_{0} I}{2 π a}$

3 $\frac{\sqrt{2} μ_{0} I}{3 π a}$

4 $\frac{2 μ_{0} I}{3 π a}$

Ans.
(3)

$\vec{B} = {\vec{B}}_{A B} + {\vec{B}}_{B C} + {\vec{B}}_{C D} + {\vec{B}}_{D A}$

${\vec{B}}_{A B} = {\vec{B}}_{B C}$

   $= \frac{μ_{0} (2 I / 3)}{4 π (a / 2)} [\sin 45 ° + \sin 45 °] (- \hat{k})$

   $= \frac{μ_{0} I \sqrt{2}}{3 π a} (- \hat{k})$

${\vec{B}}_{C D} = {\vec{B}}_{D A}$

$= \frac{μ_{0} (I / 3)}{4 π (a / 2)} [\sin 45 ° + \sin 45 °] (\hat{k})$

   $= \frac{μ_{0} I \sqrt{2}}{6 π a} (\hat{k})$

$\vec{B} = 2 [\frac{μ_{0} I \sqrt{2}}{3 π a} (- \hat{k}) + \frac{μ_{0} I \sqrt{2}}{6 π a} (\hat{k})]$

$\Rightarrow \vec{B} = \frac{\sqrt{2} μ_{0} I}{3 π a} (- \hat{k})$

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