Question

(hrw8c29p68) Each of two long straight parallel wires 13 cm apart carries a current of 150 A. The figure shows a cross section, with the wires running perpendicular to the page and point P lying on the perpendicular bisector of the line between the wires. Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is also out of the page. AZ Wire 1 Wire 2 What is the magnitude of B? 4.62x 10-41 You are correct. Previous Tries Your receipt no. is 168-8722 What is the value of Bx? 0.00T You are correct. Previous Tries Your receipt no. is 168-5402 What is the value of By? -3.26x10^-41 Submit Answer Incorrect. Tries 6/10 Previous Tries What is the value of Bz? 0.00 T You are correct. Previous Tries Your receipt no. is 168-6839 Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is into the page. What is the magnitude of B? Submit Answer Tries 0/10 What is the value of Bx? 0.00T You are correct. Your receipt no. is 168-79220 What is the value of By? 0.00T Previous Tries You are correct. Provou Trier

Answer 1

Write the expression to determine the magnetic field. B-ul 24 (41x10->)x150 27x0.13 = 3.264x10+1 From wire 1 and wire 2 B = B(-cos 45j +sin 45k) B, = B(-cos 459-sin 45£) Thus net magnetic field at Pis B = 3, +3, B =-2B cos 459 =-2(3.264x10* T)cos 45ị 4.62x10^T Therefore, B = 0 B, =-4.62x10* Tj B.=0 = B = 4.62x1041

(b) From wire 1 and wire 2 B = B(-cos 45j +sin 454) B2 =B cos 459 +sin 45€) Thus net magnetic field at Pis B = 3, +B, B = 2B sin 45k = 2(3.264x10-4 T )sin 45ł = 4.62x104 TÂ Therefore, B = 0 B=0 B. = 4.62x1041 B = 4.62x1041