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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 . |
| Answer» <html><body><p></p>Solution : Length of copper <a href="https://interviewquestions.tuteehub.com/tag/wire-1457703" style="font-weight:bold;" target="_blank" title="Click to know more about WIRE">WIRE</a>, `l = 0.5` m <br/>Area of cross section, `A = 10 mm^(2) = 10 xx 10^(-6) m^(2)` <br/>Number density of electrons, `n = 8 xx 10^(28) m^(-3)` <br/>Total charge inside the wire is given by <br/> `<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a> = <a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a> xx ` Volume of wire <br/> `= ne xx Al` <br/> ` = 8 xx 10^(28) xx 1.6 xx 10^(-19) xx 10 xx 10^(-6) xx 0.5` <br/> ` = 6.4 xx 10^(4) C` <br/> Let `v_(d)` be the drift <a href="https://interviewquestions.tuteehub.com/tag/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> of electrons. Then, magnetic force on the wire due to external magnetic field B will be <br/> ` F = q v _(D) B sin theta` <br/> Here, `theta = 90^(@) , B = 10 " gauss " = 10 xx 10^(-4) T = 10^(-3) T`<br/> ` rArr "" F = qv_(D) B sin theta` <br/> ` rArr "" 0.06 = 6.4 xx 10^(4) xx v_(D) xx 10^(-3) xx 1` <br/> ` rArr "" v_(D) = (0.06)/(6.4 xx 10^(4) xx 10^(-3)) = 9.4 xx 10^(-4) ` m/s</body></html> | |