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A ciruclar coil of 20 turns and radius `10cm` is placed in a uniform magnetic field of `0.1T` normal to the plane of the coil . If the current in the coil is `5.0A` what is the average force on each electron in the coil due to the magnetic field `(` The coil is made of copper wire of cross`-` sectional area `10^(-5)m^(2)` and the free electron density in copper is given to be about `10^(29)m^(-3))`. |
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Answer» n=20, r=10 cm, B=0.10 T, I=5.0 A Force on each electron n=No. of electrons per unit volume A= Area of cross-section of wire `F=Be V=(BI)/(nA)=(0.1xx5)/(10^(29)xx10^(-5))=5xx10^(-25)N` |
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