Answer:
The force that the wire exerts on the electron is [tex]-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Explanation:
Given that,
Current = 8.60 A
Velocity of electron [tex]v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s[/tex]
Position of electron = (0,0.200,0)
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B=\dfrac{\mu I}{2\pi d}(-k)[/tex]
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}[/tex]
[tex]B=0.0000086\ T[/tex]
[tex]B=-8.6\times10^{-6}k\ T[/tex]
We need to calculate the force that the wire exerts on the electron
Using formula of force
[tex]F=q(\vec{v}\times\vec{B}[/tex]
[tex]F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )[/tex]
[tex]F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k[/tex]
[tex]F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Hence, The force that the wire exerts on the electron is [tex]-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
