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Math Help - Average force on an electron in a wire

  1. #1
    Member alexgeek's Avatar
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    Question 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 8.0\, A is passed through a conductor of length 0.40 \,m and cross-sectional area 1.0\times 10^{-6} \:m^2. The conductor contains 8.0\times 10^{28} electrons per m^3. When the conductor is at right angles to a magnetic field of flux density 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:
    <br />
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:

    <br />
\frac{F}{N_e} = \frac{0.64}{8.0 \times 10^{28}} = 8.0 \times 10^{-30} \, N

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

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

    Thanks.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    conductor volume ... V = (0.4)(10^{-6}) = 4 \times 10^{-7} \, m^3

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

    \frac{F}{N_e} = \frac{0.64}{3.2 \times 10^{22}} = 2 \times 10^{-23}
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  3. #3
    MHF Contributor Unknown008's Avatar
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    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.
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