Part A Learning Goal: To understand the magnetic force on a straight current-carrying wire in a uniform magnetic field. Consider a wire of length L = 0.30 m that runs north-south on a horizontal surface. There is a current of 1 = 0.50 A flowing north in the wire. (The rest of the circuit, which actually delivers Magnetic fields exert forces on moving charged particles, whether those this current, is not shown.) The Earth's magnetic field at this location has a magnitude of 0.50 gauss (or, in SI units, 0.5 x 10-4 tesla) and points north and 38 degrees down from the charges are moving independently or are confined to a current-carrying wire. horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that 1 T = 1 N/(A . m).( Figure 1) The magnetic force F on an individual moving charged particle depends on its velocity v and charge q. In the case of a current-carrying wire, many charged Express your answer in newtons to two significant figures. particles are simultaneously in motion, so the magnetic force depends on the total current I and the length of the wire L. View Available Hint(s) The size of the magnetic force on a straight wire of length L carrying current I in a uniform magnetic field with strength B is IVO AEQ ? F = ILB sin(0) Here O is the angle between the direction of the current (along the wire) and 5.91 . 10-5 N the direction of the magnetic field. Hence B sin(0) refers to the component of the magnetic field that is perpendicular to the wire, BJ . Thus this equation can also be written as F = ILBI . Submit Previous Answers The direction of the magnetic force on the wire can be described using a "right- hand rule." This will be discussed after Part B. X Incorrect; Try Again; 5 attempts remaining Part B Figure