Question

A length L of wire carrying current I is bent into a circle of one turn. The field at the center of the coil is B₁. A similar wire of length L carrying current I is bent into a square of one turn. The field at its center is B₂. Then-

• A. B₁> B₂
• B. B₁< B₂
• C. B₁ = B₂
• D. Nothing can be predicted

Answer 1

Answer: (B)

B₁< B₂

Answer 2

Magnetic field due to circle

B₁ = but L = 2pR Ž 2R = $\frac { \mathrm { L } } { \pi }$

Ž B₁ = $\frac { \mu _ { 0 } \pi \mathrm { I } } { \mathrm { L } }$ $\simeq$ 3.14

Magnetic field due to square coil B₂ = $2 \sqrt { 2 } \left( \frac { \mu _ { 0 } I } { \pi x } \right)$

but L = 4x Ž x = $\frac { L } { 4 }$

\ B₂ = $8 \sqrt { 2 } \frac { \mu _ { 0 } \mathrm { I } } { \pi \mathrm { L } }$ $\simeq$ 3.60

\ B₁ < B₂