Question

The current loop ABCDA carries current 37.2 A in the direction indicated in the figure, where the segments AB and CD are concentric arcs with radii 7 cm and 11 cm respectively. What is the magnitude of the resultant magnetic field ||B|| at the center of curvature O due to the current segment of arc CD? Answer in units of T.

Answer 1

Magnetic field due to segment of a circle at the center is given by:

B = (μ₀I/2r)*(θ/360) where I is the current, r is the radius of circle and θ is the arc angle subtended at the center

Now we have two different arcs so:

magnetic field due to inner arc: B₁  =   (μ₀I/2r₁)*(θ/360) (going into the paper)

magnetic field due to outer arc: B₂   = (μ₀I/2r₂)*(θ/360) (coming out of the paper)

Hence the net magnetic field at the center B = (μ₀Iθ/720)*( 1/r₁ - 1/r₂) (going into the paper)

There will be no contribution for the straight wires as they lie along the center.

Now we are given: I = 37.2 A; r₁ = 7 cm; r₂  = 11 cm; and   θ = 75⁰

Hence B = (4π*10^-7*37.2*75/720)*( 1/7 - 1/11)*100

B = 2.53*10^-5 T