Practice portal › Moving Charges and Magnetism › Biot-Savart Law: Straight Wires and Arcs

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 B_A and B_B are the values of magnetic field at the centres of the circle and square respectively, then the ratio (B_A)/(B_B) is

Asked in JEE Main 2016 · Arcs and combined shapes

Answer: (4) (π²)/(8√2)

Step-by-step solution

Circle: R=l/(2π), B_A=(μ₀ I)/(2R)=(πμ₀ I)/l.

Square: a=l/4, B_B=4×(μ₀ I)/(4π(a/2))(2 sin 45°)=(2√2μ₀ I)/(π a)=(8√2μ₀ I)/(π l).

(B_A)/(B_B)=(π²)/(8√2)

Why the other options are wrong

More Biot-Savart Law: Straight Wires and Arcs questionsAll Biot-Savart Law: Straight Wires and Arcs questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer