Question

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

• A. $\frac { \sqrt { 2 } \mu _ { 0 } i } { 3 \pi a }$◉
• B. $\frac { \sqrt { 2 } \mu _ { 0 } \mathrm { i } } { 3 \pi \mathrm { a } } \otimes$
• C. $\frac { \sqrt { 2 } \mu _ { 0 } \mathrm { i } } { \pi \mathrm { a } }$◉
• D. $\frac { \sqrt { 2 } \mu _ { 0 } \mathrm { i } } { \pi \mathrm { a } } \otimes$

Answer 1

Answer: (B)

$\frac { \sqrt { 2 } \mu _ { 0 } \mathrm { i } } { 3 \pi \mathrm { a } } \otimes$

Answer 2

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 $i _ { 1 } + i _ { 2 } = i$ ⇒ $i _ { 1 } = \frac { 2 i } { 3 }$ and $i _ { 2 } = \frac { i } { 3 }$

Magnetic field at centre O due to wire AB and BC (part 1 and 2) $B _ { 1 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 i _ { 1 } \sin 45 ^ { \circ } } { a / 2 } \otimes$ $= \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \sqrt { 2 } i _ { 1 } } { a } \otimes$

and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) $B _ { 3 } = B _ { 4 } = \frac { \mu _ { 0 } } { 4 \pi } \frac { 2 \sqrt { 2 } i _ { 2 } } { a }$◉

Also i₁ = 2i₂. So (B₁ = B₂) > (B₃ = B₄)

Hence net magnetic field at centre O

$B _ { \text {net } } = \left( B _ { 1 } + B _ { 2 } \right) - \left( B _ { 3 } + B _ { 4 } \right)$ $= 2 \times \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \sqrt { 2 } \times \left( \frac { 2 } { 3 } i \right) } { a } - \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \sqrt { 2 } \left( \frac { i } { 3 } \right) \times 2 } { a }$