Question

A long straight wire carries a current of 51.7 A. An electron, traveling at 7.43 × 10^7 m/s, is 5.76 cm from the wire.

Answer 1

Answer:

(a) 2.13*10^{-5} N

(b) 2.13*10^{-5} N

(c) 0 N

Explanation:

the magnitude of the magnetic field generated by the wire is:

$\frac{\mu_0I}{2\pi r}=\frac{(4\pi*10^{-7})(51.7A)}{2\pi(5.76*10^{-2}m)}=1.79*10^{-4}T$

if we assume that the current is in the +y direction, B is in the +z direction.

(a) toward the wire, electron is in the -x direction. The angle between B and v is 90°. By using the following formula we obtain:

$F_1=qvBsin90\°=(1.6*10^{-19}C)(7.43*10^{7}\frac{m}{s})(1.79*10^{-4}T)=2.13*10^{-15}N$

(b) parallel to the wire, electron is in the +y direction. Again angle between B ans v is 90°.

$F_2=qvB=2.13*10^{-15}N$

(c) perpendicular to both previous directions, that is, +z or -z. In this case velocity vector is parallel to the magnetic field vector. Hence:

F3=0N

hope this helps!