Two identical wires A and B, each of length 'l', carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side 'a'. If BA and BB are the values of magnetic fields at the centres of the circle and square respectively, then the ratio (BA)/(BB) is: (a) (π²)/(16) (b) (π²)/(8√(2)) (c) (π²)/(8) (d) (π²)/(16√(2))

Answer: (b). Circle: 2π R = l ⇒ BA = (μ₀ I)/(2R) = (πμ₀ I)/(l). Square: a = l/4, BB = (2√(2)μ₀ I)/(π a) = (8√(2)μ₀ I)/(π l). (BA)/(BB) = (π²)/(8√(2)) So the answer is (b).

Two identical wires A and B, each of length 'l', carry the same… | JEE Main 2016

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Magnetic Field of Straight Wires

Two identical wires $A$ and $B,$ each of length $\text{'}l\text{'}$ , carry the same current $I$ . Wire $A$ is bent into a circle of radius $R$ and wire $B$ is bent to form a square of side $\text{'}a\text{'}$ . If $B_A$ and $B_B$ are the values of magnetic fields at the centres of the circle and square respectively, then the ratio $\frac{B_A}{B_B}$ is:

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Answer: (b)

Circle: $2\pi R = l \Rightarrow B_A = \frac{\mu_0 I}{2R} = \frac{\pi\mu_0 I}{l}$ .
Square: $a = \frac{l}{4}$ , $B_B = \frac{2\sqrt{2}\mu_0 I}{\pi a} = \frac{8\sqrt{2}\mu_0 I}{\pi l}$ .
$\frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt{2}}$
So the answer is (b).

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