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InterviewSolution
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A copper wire of length 0.5 m and cross sectional area 10mm^(2) , carries a current perpendicular to a magnetic field of strength 10gauss. As a result, the wire experiences a force of 0.06 N. What is the drift velocity of free electrons in the copper wire? Take the number density of electrons= 8 xx 10^(28) m^(-3) and atomie weight of copper= 63.5 . |
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Answer» Solution : Length of copper WIRE, `l = 0.5` m Area of cross section, `A = 10 mm^(2) = 10 xx 10^(-6) m^(2)` Number density of electrons, `n = 8 xx 10^(28) m^(-3)` Total charge inside the wire is given by `Q = NE xx ` Volume of wire `= ne xx Al` ` = 8 xx 10^(28) xx 1.6 xx 10^(-19) xx 10 xx 10^(-6) xx 0.5` ` = 6.4 xx 10^(4) C` Let `v_(d)` be the drift SPEED of electrons. Then, magnetic force on the wire due to external magnetic field B will be ` F = q v _(D) B sin theta` Here, `theta = 90^(@) , B = 10 " gauss " = 10 xx 10^(-4) T = 10^(-3) T` ` rArr "" F = qv_(D) B sin theta` ` rArr "" 0.06 = 6.4 xx 10^(4) xx v_(D) xx 10^(-3) xx 1` ` rArr "" v_(D) = (0.06)/(6.4 xx 10^(4) xx 10^(-3)) = 9.4 xx 10^(-4) ` m/s |
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