Question

An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.60 A in the -c-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron's velocity is v(5.00 What is the force that the wire exerts on the electron? Enter the z, y, and z components of the force separated by commas. 104 m/s)^-(3.00 x 104 m/s)3.

Answer 1

Answer:

The  force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Explanation:

Given that,

Current = 8.60 A

Velocity of electron $v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s$

Position of electron = (0,0.200,0)

We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{\mu I}{2\pi d}(-k)$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}$

$B=0.0000086\ T$

$B=-8.6\times10^{-6}k\ T$

We need to calculate the force that the wire exerts on the electron

Using formula of force

$F=q(\vec{v}\times\vec{B}$

$F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )$

$F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k$

$F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Hence, The  force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$