Figure shows a current carrying square loop ABCD of edge length is a lying in a plane. If the resistance of th…

Question

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 the ADC part is $2r$ , then the magnitude of the resultant magnetic field at the center of the square loop is:

A square loop ABCD drawn like a diamond, with current entering at A and leaving at C, currents i1 through path ABC and i2 through path ADC, edge length a marked on side BC, and center marked by a dot.

A
$\frac{\sqrt{2}\,\mu_0 I}{3 \pi a}$
B
$\frac{\mu_0 I}{2 \pi a}$
C
$\frac{2 \mu_0 I}{3 \pi a}$
D
$\frac{3 \pi \mu_0 I}{\sqrt{2}\,a}$

Accepted Answer

A. $$\frac{\sqrt{2}\,\mu_0 I}{3 \pi a}$$

Answer

B. $$\frac{\mu_0 I}{2 \pi a}$$

Answer

C. $$\frac{2 \mu_0 I}{3 \pi a}$$

Answer

D. $$\frac{3 \pi \mu_0 I}{\sqrt{2}\,a}$$

Answer & step-by-step solution

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