Answer:
The magnitude of the force on the moving charge is [tex]3\times10^{-9}\ N[/tex]
Explanation:
Given that,
Current = 5.0 A
Charge [tex]q=-3\times10^{-6}\ C[/tex]
Speed v = 100 m/s
Distance = 0.1
We need to calculate the magnitude of the force on the moving charge
Using formula of force
[tex]F=qvB[/tex]
[tex]F=qv\times\dfrac{\mu_{0}I}{2\pi r}[/tex]
Put the value into the formula
[tex]F=-3\times10^{-6}\times100\times\dfrac{4\pi\times10^{-7}\times5.0}{2\pi\times0.1}[/tex]
[tex]F=-3\times10^{-9}\ N[/tex]
Hence, The magnitude of the force on the moving charge is [tex]3\times10^{-9}\ N[/tex]
