A wire A, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm and another wire B, also bent in the shape of arc of a circle, carrying a current of 3 A and having radius of 4 cm, are placed as shown in the figure. The ratio of the magnetic fields due to the wire…

Question

A wire A, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm and another wire B, also bent in the shape of arc of a circle, carrying a current of 3 A and having radius of 4 cm, are placed as shown in the figure. The ratio of the magnetic fields due to the wires A and B at the common centre O is

Accepted Answer

(1) 6:5. Arc field B=(μ₀ I)/(2R)·θ/(2π); the radial straight parts give zero. A: gap 90°, arc 270°: B_A∝2/(0.02)·3/4=75 B: gap 60°, arc 300°: B_B∝3/(0.04)·5/6=62.5 B_A:B_B=75:62.5=6:5

Step-by-step solution

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