Question

A very long, straight wire carries a current of 19.0 A in the +k direction. An electron 1.9 cm from the center of the wire in the î direction is moving with a speed of 3.23 106 m/s. Find the force on the electron when it moves in the following directions. (a) directly away from the wire 0 Correct: Your answer is correct. N î + 0 Correct: Your answer is correct. N ĵ + N k (b) parallel to the wire in the direction of the current N î + 0 Correct: Your answer is correct. N ĵ + 0 Correct: Your answer is correct. N k (c) perpendicular to the wire and tangent to a circle around the wire in the + ĵ direction 0 Correct: Your answer is correct. N î + 0 Correct: Your answer is correct. N ĵ + N k

Answer 1

(a) $1.03\cdot 10^{-16} N$, -k direction

First of all, let's find the magnetic field produced by the wire at the location of the electron:

$B=\frac{\mu_0 I}{2 \pi r}$

where

I = 19.0 A is the current in the wire

r = 1.9 cm = 0.019 m is the distance of the electron from the wire

Substituting,

$B=\frac{(1.256\cdot 10^{-6})(19.0A)}{2 \pi (0.019 m)}=2\cdot 10^{-4} T$

and the direction is +j direction (tangent to a circle around the wire)

Now we can find the force on the electron by using:

$F=qvBsin \theta$

where

$q=1.6\cdot 10^{-19}C$ is the electron's charge

$v=3.23\cdot 10^6 m/s$ is the electron speed

$B=2\cdot 10^{-4} T$ is the magnetic field

$\theta$ is the angle between the direction of v and B

In this case, the electron is travelling away from the wire, while the magnetic field lines (B) form circular paths around the wire: this means that v and B are perpendicular, so $\theta=90^{\circ}, sin \theta=1$. So, the force on the electron is

$F=(1.6\cdot 10^{-19}C)(3.23\cdot 10^6 m/s)(2\cdot 10^{-4} T)(1)=1.03\cdot 10^{-16} N$

The direction is given by the right hand rule:

- Index finger: direction of motion of the electron, +i direction (away from the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means -k direction (anti-parallel to the current in the wire)

(b) $1.03\cdot 10^{-16} N$, +i direction

The calculation of the magnetic field and of the force on the electron are exactly identical as before. The only thing that changes this time is the direction of the force. In fact we have:

- Index finger: direction of motion of the electron, +k direction (parallel to the current in the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means +i direction (away from the wire)

(c) 0

In this case, the electron is moving tangent to a circle around the wire, in the +j direction. But this is exactly the same direction of the magnetic field: this means that v and B are parallel, so $\theta=0, sin \theta=0$, therefore the force on the electron is zero.