## Related questions with answers

A strip of copper is placed in a uniform magnetic field of magnitude 2.5 T. The Hall electric field is measured to be $1.5 \times 10 ^ { - 3 } \mathrm { V } / \mathrm { m }$. (a) What is the drift speed of the conduction electrons? (b) Assuming that $\mathrm { n } = 8.0 \times 10 ^ { 28 }$ electrons per cubic meter and that the cross-sectional area of the strip is $5.0 \times 10 ^ { - 6 } \mathrm { m } ^ { 2 }$, calculate the current in the strip. (c) What is the Hall coefficient 1/nq?

Solution

Verifieda)In studying the Hall effect, equilibrium is reached when the magnitude of the magnetic force on the moving charges equal the magnitude of the electrical force caused by the electric field created by the separated charges:

$qE=qv_dB$

Solving for the drift velocity of the charges $v_d$:

$v_d=\dfrac{E}{B}=\dfrac{1.5 \times 10^{-3}}{2.5}=6 \times 10^{-4} \, \mathrm{m/s}$

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