An infinite wire has a circular bend of radius a , and carrying a current I as shown in the figure. The magnitude of magnetic field at the origin O of the arc is given by:

Solution

Physics sir · 3 steps

Split the wire into three pieces, find each piece's field at O, and add the ones that survive.

1. Upper straight wire. This wire lies along the line through O. For every element, dl⃗ is parallel to r⃗, so dl⃗ × r⃗ = 0 and its field at O is zero.

   Piece	Field at O	Direction
   Upper straight wire (line passes through O)	0	none
   Arc of angle 3π/2 (three-quarter circle)	(μ₀ I)/(4π a) × 3π/2	into the page (clockwise current)
   Lower wire, semi-infinite, ends at the foot of the perpendicular from O	(μ₀ I)/(4π a)	into the page

2. Check the two pieces. The arc runs from the bottom, round the left side and over the top to the right: that is 3π/2 of a circle, and an arc of angle θ gives (μ₀ I θ)/(4π a). The lower wire starts exactly below O, so it is half of an infinite wire and gives half of (μ₀ I)/(2π a). By the right-hand rule, both fields point into the page, so they add.

   B = (μ₀ I)/(4π a)·3π/2 + (μ₀ I)/(4π a) = μ₀/4πI/a[3π/2 + 1]

3. So. The magnitude of the field at O is μ₀/4πI/a[3π/2+1], which is option (b).

The idea to keep. A wire whose line passes through the point gives zero field. A wire that ends at the foot of the perpendicular gives half the infinite-wire value. An arc gives (μ₀ I θ)/(4π a).

Answer from the NTA final answer key. Solution by PrepWiser.