F on an individual moving particle depends on its velocity v and charge q. In th
ID: 1420263 • Letter: F
Question
F on an individual moving particle depends on its velocity v and charge q. In the case of a current-carrying wire, many charged particles are simultaneously in motion, so the magnetic force depends on the total current I and the length of the wire L. The size of the magnetic force on a straight wire of length L carrying current I in a uniform magnetic fie with strength B is F= ILBsm(theta). Here theta is the angle between the direction of the current (along the wire) and the direction of the magnetic field. Hence Bsin(theta) refers to the component of the magnetic field that is perpendicular to the wire, B. Thus this equation can also be written as F = ILB. Consider a wire of length L = 0.30 m that runs north-south on a horizontal surface. There is a current of I = 0.50 A flowing north in the wire. (The rest of the circuit, which actually delivers 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 times 10^-4 tesla) and points north and 38 degrees down from the 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) Express your answer in newtons to two significant figures. Now assume that a strong, uniform magnetic field of size 0.55 T pointing straight down is applied. What is the size of the magnetic force on the wire due to this applied magnetic field? Ignore the effect of the Earth's magnetic field. Express your answer in newtons to two significant figures. The direction of the magnetic force is perpendicular to both the direction of the current flow and the direction of the magnetic field. Here is a "right-hand rule" to help you determine the direction of the magnetic force. (Figure 2) 1. Straighten the fingers of your right hand and point them in the direction of the current. 2. Rotate your arm until you can bend your fingers to point in the direction of the magnetic field.Explanation / Answer
i = east
-i west
j = north
-j south
k up
-k dowm
part A
F = I*(LxB)
L = 0.3j
B = (0.5*10^-4*cos38j) - (0.5*10^-4*sin38)k
F = 0.5*0.3*0.5*10^-4*sin38 (-i)
F = 4.62*10^-6 (-i)
west
partB)
L = 0.3j
B = (0.55*10^-4)(-k)
F = 0.5*0.3*0.55*10^-4 (-i)
F = 8.25*10^-6 (-i)
direction west
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