Question

A long, straight wire carries a current of 8.60
a. an electron is traveling in the vicinity of the wire. part a at the instant when the electron is 3.90 cm from the wire and traveling with a speed of 5.00×104 m/s directly toward the wire, what is the magnitude of the force that the magnetic field of the current exerts on the electron?

Answer 1

First of all, we need to calculate the magnetic field generated by the wire at a distance r=3.90 cm=0.039 m, which is given by:

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

where I=8.60 A is the current. Substituting numbers in the equation, we find

$B=\frac{(4 \pi \cdot 10^{-7})(8.6 A)}{2 \pi (0.039 m)}=4.4 \cdot 10^{-5} T$

And now we can calculate the force exerted on the electron, which is given by:

$F=qvB$

where $q=1.6 \cdot 10^{-19} C$ is the charge of the electron and $v=5.0 \cdot 10^4 m/s$ is its speed. Substituting data in the formula, we find

$F=(1.6 \cdot 10^{-19}C)(5.00 \cdot 10^4 m/s)(4.4 \cdot 10^{-5} T)=3.5 \cdot 10^{-19} N$