Question

Learning Goal: To practice Problem-solving Strategy 24.2 Magnetic force problems A long straight horizontal wire carries a current I = 3.70 A to the left. A positive 1.00 C charge moves to the right at a distance 1.50 m above the wire at constant speed v = 3250 m/s. (Figure 1) What are the magnitude and the Figure < 1 of 1 1 Copyright 2020 Pearson Education !

Answer 1

magnetic field due to current-carrying wire at the location of charge is,

$B= \frac{\mu_0 I}{2 \pi r}$

= (2 x 10^-7) x 3.70 / 1.50

= 4.93 x 10^-7 T , directed into the page

( note that $\frac{\mu_ 0}{2 \pi } =$ 2 x 10^-7 )

magnetic force on the charge is given by,

= 1.00 x 3250 x (4.93 x 10^-7 )

= 1.60 x 10^-3 N