Answer:
[tex]2.34\cdot 10^{-19}N[/tex]
Explanation:
The strength of the magnetic field produced by a current-carrying wire is:
[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]
where:
[tex]\mu_0[/tex] is the vacuum permeability
I is the current in the wire
r is the distance from the wire
In this case,
I = 5.20 A
r = 4.40 cm = 0.044 m
Therefore,
[tex]B=\frac{(4\pi \cdot 10^{-7})(5.20)}{2\pi (0.044)}=2.36\cdot 10^{-5} T[/tex]
The direction of the field is along concentric circles centered in the wire.
The force exerted by a magnetic field on a moving charged particle is
[tex]F=qvB sin \theta[/tex]
where
q is the charge
v is the velocity of the particle
[tex]\theta[/tex] is the direction between v and B
In this problem:
[tex]q=1.6\cdot 10^{-19}C[/tex] is the magnitude of the charge of the particle
[tex]v=6.20\cdot 10^4 m/s[/tex] is the velocity
[tex]\theta=90^{\circ}[/tex], because the electron is travelling towards the wire, so perpendicular to the lines of the magnetic field
Therefore, the magnitude of the force is:
[tex]F=(1.6\cdot 10^{-19})(2.36\cdot 10^{-5})(6.20\cdot 10^4)(sin 90^{\circ})=2.34\cdot 10^{-19}N[/tex]
