Figure shows a square loop ABCD with edge length a. The resistance of the wire A

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Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. The value of magnetic field at the centre of the loop assuming uniform wire is

Figure Shows A Square Loop ABCD With Edge Length A. The Resi...

Options:

A . √2μ0i3πa⊙

B . √2μ0i3πa⊗

C . √2μ0iπa⊙

D . √2μ0iπa⊗

Answer: Option B
:
B
According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is half that of ABC i.e.

\frac{i_{2}}{i_{1}} = \frac{1}{2}

. Also

i_{1} + i_{2} = i \Rightarrow i_{1} = \frac{2 i}{3}

and

i_{2} = \frac{i}{3}

Magnetic field at centre O due to wire AB and BC (part 1 and 2)

B_{1} = B_{2} = \frac{μ_{0}}{4 π} . \frac{2 i_{1} s i n 45^{\circ}}{a / 2} \otimes = \frac{μ_{0}}{4 π} . \frac{2 \sqrt{2} i_{1}}{a} \otimes

and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4)

B_{3} = B_{4} = \frac{μ_{0}}{4 π} \frac{2 \sqrt{2} I_{2}}{a} ⊙

Also

i_{1} = 2 i_{2}

. So

(B_{1} = B_{2}) > (B_{3} = B_{4})

Hence net magnetic field at centre O

B_{n e t} = (B_{1} + B_{2}) - (B_{3} + B_{4})

= 2 \times \frac{μ_{0}}{4 π} . \frac{2 \sqrt{2} \times (\frac{2}{3} i)}{a} - \frac{μ_{0}}{4 π} . \frac{2 \sqrt{2} (\frac{i}{3}) \times 2}{a}

= \frac{μ_{0}}{4 π} . \frac{4 \sqrt{2} i}{3 a} (2 - 1) \otimes = \frac{\sqrt{2} μ_{0} i}{3 π a} \otimes

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