can you explain the solution?An infinitely long wire is given a hairpin bend as shown. Point a is at the centre of the circle creating the hairpin bend in the wire. Point b is equal distance from both wires R = 25 mm and is far from the curve such that the field produced by the curve can be ignored. A current of 16 A runs through the wire. (a) What is the magnitude and direction of the magnetic field at point a? (b) What is the magnitude and direction of the magnetic field at point b? Solution (a) At Point a, there are only "long" wires to the right. This would give the a field that is half the field from a long wire, so B 1 2 wire = 1 2 Bwire = 1 2 ( I 2R ) = I 4R Similarly, the arc is a half-circle, and the B-field will be half of the field from a full circle, giving B 1 2 circle = 1 2 Bcircle = 1 2 ( I 2R ) = I 4R By the right-hand rule, the direction of all contributions to the magnetic field at Point a are the smallout of the page. So, the total field is found by adding the contributions of the two half-wires and the half-circle. Btotal = 2B 1 2 wire + B 1 2 circle Btotal = 2 ( I 4R ) + I 4R = I R ( 1 2 + 1 4 ) Btotal = (4 107)(16) 25 103 ( 1 2 + 1 4 ) Btotal = 3.29 104 T