Question

At the instant when the electron is 4.40 cmcm from the wire and traveling with a speed of 6.10×104 m/sm/s directly toward the wire, what is the magnitude of the force that the magnetic field of the current exerts on the electron?

Answer 1

Answer:

The force that the magnetic field of the current exerts on the electron is $2.30\times10^{-19}\ N$

Explanation:

Given that,

Distance = 4.40 cm

Speed $v= 6.10\times10^{4}\ m/s$

Suppose a long, straight wire carries a current of 5.20 . An electron is traveling in the vicinity of the wire.

We need to calculate the magnetic field of the current exerts on the electron

Using formula of force

$F=qv\times B$

$F=qv\times\dfrac{\mu_{0}I}{2\pi r}$

Put the value into the formula

$F=1.6\times10^{-19}\times6.10\times10^{4}\times\dfrac{4\pi\times10^{-7}\times5.20}{2\pi\times4.40\times10^{-2}}$

$F=2.30\times10^{-19}\ N$

Hence, The force that the magnetic field of the current exerts on the electron is $2.30\times10^{-19}\ N$