1. * Example 29.1 A person scuffing their feet has picked up 85 mC of charge. As they walk at1.5 m/s, what is the magnetic field they induce 1 m ahead of them and 1 m to the right of thems?

EXAMPLE 29.1 | The magnetic field of a proton A proton moves with velocity v = 1.0 X 10 7 m/s. As it passes the SOLVE Position 1, which is along the line of motion, has d = 0". origin, what is the magnetic field at the (x, y, z) positions (1 mm, Thus B = 0. Position 2 (at ( mm, 1 mm, ( mm) is at distance 0 mm, 0 mm), (0 mm. 1 mm. ( mm), and (1 mm, 1 mm, 0 mm)? ", = 1 mm = 0.001 m. Equation 29.1, the Biot-Savart law, gives us MODEL The magnetic field is that of a moving charged particle. the magnetic field strength at this point as VISUALIZE FIGURE 29.9 shows the geometry. The first point is on An qusine, B = the x-axis, directly in front of the proton. with , = 0". The second 4 71 point is on the y-axis, with @2 - 90%, and the third is in the xy-plane. 47 X 10"'Tm/A (1.60 X 10""'C) (1.0 * 10' m/s ) sin 90" 4 1 (0.0010 m)= FIGURE 29.9 The magnetic field of Example 29.1. = 1.60 X 10-13 T Position 3 According to the right-hand rule, the field points in the positive z-direction. Thus Position 2 B, = 1.60 X 10-13 X T 4 where k is the unit vector in the positive z-direction. The field at position 3, at (1 mm, 1 mm, 0 mm), also points in the z-direction, but Position I it is weaker than at position 2 both because r is larger and because e B= 0 is smaller. From geometry we know ry = V2 mm = 0.00141 m and 0; - 45". Another calculation using Equation 29.1 gives B, = 0.57 X 10-13 & T