Description: Exercise 28.42 A hairpin configuration is formed of two semi-infini
ID: 2298202 • Letter: D
Question
Description: Exercise 28.42
A hairpin configuration is formed of two semi-infinite straight wires that are 5.00cm apart and joined by a semicircular piece of wire (whose radius must be 2.50cm and whose center is at the origin of xyz-coordinates). The top wire lies along the line y=2.50cm, and the bottom wire lies along the line y=?2.50cm; these two wires are in the left side (x<0) of the xy-plane. The current in the hairpin is 5.50A, and it is directed toward the right in the top wire, clockwise around the semicircle, and to the left in the bottom wire. Find the magnetic field at the origin of the coordinate system.
Please include a multiplication sign (*) to indicate when two units are multiplied together.
B?net=[Num] [Units]z^
A hairpin configuration is formed of two semi-infinite straight wires that are 5.00cm apart and joined by a semicircular piece of wire (whose radius must be 2.50cm and whose center is at the origin of xyz-coordinates). The top wire lies along the line y=2.50cm, and the bottom wire lies along the line y=?2.50cm; these two wires are in the left side (x<0) of the xy-plane. The current in the hairpin is 5.50A, and it is directed toward the right in the top wire, clockwise around the semicircle, and to the left in the bottom wire. Find the magnetic field at the origin of the coordinate system.
Please include a multiplication sign (*) to indicate when two units are multiplied together.
B?net=[Num] [Units]z^
Explanation / Answer
The field due to a semi infinite wire = mu I / 4 pi R
mu = 4pi * 10-7, I = 5.50 A, R = 2.50 cm
Field due to semi circular ring mu I / 4 R
substituting the values we get B
B = 4pi * 10-7 * 5.50 / 4 * 2.50* 10-2
= 6.91 * 10-5 T