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Copper has 8.0xx 10^(28) electrons per cuble meter carrying a current and lying at right angle to a magnetic field of strength 5 xx 10^(-3) T. experiences a force of 8.0 xx 10^(-2) N.Calculate the drift velocity of free electrons in the wire. |
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Answer» Solution :n = 8 `xx 10^(28) m^(-3), l = 1m ` A = 8 `xx 10^(-6) m^(-2) , e = 1.6 xx 10^(-9)` C Total charge contained in the wire q = volume of wire `xx `ne = alne = `8 xx 10^(-6) xx 1 xx 8 xx 10^(28) xx 1.6 xx 10^(19)`C `= 102.4 xx 10^(3)` C If `v_(d)` is the drift speed of electrons, then F = `qv_(d) B sin 90^(@) = qv_(d) B ` `therefore v_(d) = (F)/(qB) = (8.0 xx 10^(-2))/(102.4 xx 10^(3) xx 5 xx 10^(-3) ) MS^(-1)` `v_(d) = 1.56 xx 10^(-4) ms^(-1)` |
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