# Average force on an electron in a wire

• Jan 26th 2011, 12:35 PM
alexgeek
Average force on an electron in a wire
Hey, just did a 25 question multiple choice paper and this was the only one that I got wrong. Was hoping someone could enlighten me.

Quote:

Originally Posted by AQA PHYA4/1 Jan 2010
A current of $\displaystyle 8.0\, A$ is passed through a conductor of length $\displaystyle 0.40 \,m$ and cross-sectional area $\displaystyle 1.0\times 10^{-6} \:m^2$. The conductor contains $\displaystyle 8.0\times 10^{28}$ electrons per $\displaystyle m^3$. When the conductor is at right angles to a magnetic field of flux density $\displaystyle 0.20\, T$, it experiences a magnetic force. What is the average magnetic force that acts on one of the free electrons in the wire?

I got the force acting on the conductor:
$\displaystyle F = BIl = 0.20 \times 8.0 \times 0.40 = 0.64N$

So figured since that is the force on the entire conductor, then the force on one electron would be the force divided by the number of electrons:

$\displaystyle \frac{F}{N_e} = \frac{0.64}{8.0 \times 10^{28}} = 8.0 \times 10^{-30} \, N$

But the answer is $\displaystyle 2.0 \times 10^{-23} \, N$

I know I am neglecting the cross-sectional area but not sure where it would come in.

Thanks.
• Jan 26th 2011, 03:11 PM
skeeter
conductor volume ... $\displaystyle V = (0.4)(10^{-6}) = 4 \times 10^{-7} \, m^3$

number of electrons in the conductor ... $\displaystyle N_e = (8 \times 10^{28})(4 \times 10^{-7}) = 3.2 \times 10^{22}$

$\displaystyle \frac{F}{N_e} = \frac{0.64}{3.2 \times 10^{22}} = 2 \times 10^{-23}$
• Jan 26th 2011, 09:25 PM
Unknown008
Yep, you need to read well your question. It says '8.0x10^28 electrons per m^3', hence you need to find the volume and the number of electrons as Skeeter showed.