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A copper has 8.0 \times 10^{38} electrons per cubic meter. A copper wire of length 3 \mathrm{~m} and cross-sectional area 9.0 \times 10^{-7} \mathrm{~m}^2 carring a current and lying at right angle to a magnetic field of strength 5 \times 10^{-4} \mathrm{~T} experiences a force of 9.2 \times 10^{-3} \mathrm{~N}.Calculate the drift velocity of free electrons in wire.

Option: 1

1.59 \times 10^{-13} \mathrm{~ms}^{-1}


Option: 2

1.29 \times 10^{-38} \mathrm{~ms}^{-1}


Option: 3

1.67 \times 10^{-12} \mathrm{~ms}^{-1}


Option: 4

2.5 \times 10^{-12} \mathrm{~ms}^{-1}


Answers (1)

best_answer

Here, 

\quad n=8.0 \times 10^{38} \quad l=3 \mathrm{~m} \begin{aligned} & A=9.0 \times 10^{-7} \mathrm{~m}^2 \quad e=1.6 \times 10^{-19} \mathrm{C} \\ & B=5 \times 10^{-4} \mathrm{~T} \quad F=9.2 \times 10^{-3} \mathrm{~N} \\ & \theta=90^{\circ} \end{aligned}

Total change contained in the wire is

q=Volume of wire \times ne

\begin{aligned} & =A ln l=\left(9.0 \times 10^{-7}\right) \times 1 \times\left(8.0 \times 10^{38}\right) \\ & \times\left(1.6 \times 10^{-19}\right) \\ & =115.2 \times 10^{-7+38+(-19)} \\ & =115.2 \times 10^{12} \\ & \end{aligned}

If v_d is the drift velocity of electrons, then

\begin{aligned} & f=q V_d B \sin \theta \\ & \Rightarrow V_d=\frac{F}{q B \sin \theta}=\frac{9.2 \times 10^{-3}}{\left(115.2 \times 10^{12}\right) \times\left(5 \times 10^{-4}\right)} \sin 90^{\circ} \\ & \end{aligned}

\begin{aligned} & =\frac{9.2}{(115.2 \times 5)} \times 10^{-3} \times 10^{-12} \times 10^4 \\ & =\frac{9.2}{576} \times 10^{-15+4} \\ & =0.0159 \times 10^{-11} \\ & =1.59 \times 10^{-13} \mathrm{~ms}^{-1} \end{aligned}

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Rishi

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