Answer:
The force is [tex]3.2\times10^{-20}\ N[/tex]
Explanation:
Given that,
A long, straight wire carries a current.
Suppose, a long, straight wire carries a current 0.86 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.50 cm from the wire and traveling at speed of [tex]6\times10^{4}\ m/s[/tex] directly toward the wire. What are the magnitude of the force that the magnetic field of the current exerts on the electron?
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B=\dfrac{\mu_{0}I}{4\pi r}[/tex]
Where, I = current
r = distance
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times0.86}{2\pi\times4.5\times10^{-2}}[/tex]
[tex]B=0.0000038\ T[/tex]
[tex]B=3.8\times10^{-6}\ T[/tex]
We need to calculate the force
Using formula of force
[tex]F=qvB\sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
Where, q = charge of electron
v = velocity of electron
B = magnetic field
Put the value into the formula
[tex]F=1.6\times10^{-19}\times6\times10^{4}\times3.8\times10^{-6}\sin90[/tex]
[tex]F= 3.2\times10^{-20}\ N[/tex]
Hence, The force is [tex]3.2\times10^{-20}\ N[/tex]