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In a metal in the solid state, such as a...

In a metal in the solid state, such as a copper wire, the atoms are strongly bound to one another and occupý fixed positions. Some electrons (called the conductor electrons) are free to move in the body of the metal while the other are strongly bound to their atoms. In good conductors, the number of free electrons is very large of the order of `10^(28)` electrons per cubic metre in copper. The free electrons are in random motion and keep colliding with atoms. At room temperature, they move with velocities of the order of `10^5` m/s. These velocities are completely random and there is not net flow of charge in any directions. If a potential difference is maintained between the ends of the metal wire (by connecting it across a battery), an electric field is set up which accelerates the free electrons: These accelerated electrons frequently collide with the atoms of the conductor, as a result, they acquire a constant speed called the drift speed which is given by `V_e =` 1/enA where I = current in the conductor due to drifting electrons, e = charge of electron, n = number of free electrons per unit volume of the conductor and A = area of cross-section of the conductor. A uniform wire of length 2.0 m and cross-sectional area `10^(-7) m^(2)` carries a current of 1.6 A. If there are `10^(28)` free electrons per m in copper, the drift speed of electrons in copper is

A

`2 mm//s`

B

`5 mm//s`

C

`10 mm//s`

D

`1 m//s`

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To find the drift speed of electrons in a copper wire, we can use the formula for drift speed: \[ V_e = \frac{I}{n \cdot e \cdot A} \] where: - \( V_e \) = drift speed of electrons (m/s) - \( I \) = current in the conductor (A) - \( n \) = number of free electrons per unit volume (m³) - \( e \) = charge of an electron (\(1.6 \times 10^{-19}\) C) - \( A \) = area of cross-section of the conductor (m²) ### Step 1: Identify the given values From the problem statement, we have: - Length of the wire = 2.0 m (not needed for this calculation) - Cross-sectional area, \( A = 10^{-7} \, \text{m}^2 \) - Current, \( I = 1.6 \, \text{A} \) - Number of free electrons per unit volume, \( n = 10^{28} \, \text{m}^{-3} \) - Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \) ### Step 2: Substitute the values into the formula Now we can substitute these values into the drift speed formula: \[ V_e = \frac{1.6}{(10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-7})} \] ### Step 3: Simplify the expression Calculating the denominator: \[ n \cdot e \cdot A = (10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-7}) = 1.6 \times 10^{2} = 160 \] Now substituting back into the formula: \[ V_e = \frac{1.6}{160} = 0.01 \, \text{m/s} \] ### Step 4: Convert to millimeters per second To convert from meters per second to millimeters per second, we multiply by 1000: \[ V_e = 0.01 \, \text{m/s} \times 1000 = 10 \, \text{mm/s} \] ### Final Answer Thus, the drift speed of electrons in the copper wire is: \[ \boxed{10 \, \text{mm/s}} \]
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In a metal in the solid state, such as a copper wire, the atoms are strongly bound to one another and occupý fixed positions. Some electrons (called the conductor electrons) are free to move in the body of the metal while the other are strongly bound to their atoms. In good conductors, the number of free electrons is very large of the order of 10^(28) electrons per cubic metre in copper. The free electrons are in random motion and keep colliding with atoms. At room temperature, they move with velocities of the order of 10^5 m/s. These velocities are completely random and there is not net flow of charge in any directions. If a potential difference is maintained between the ends of the metal wire (by connecting it across a battery), an electric field is set up which accelerates the free electrons: These accelerated electrons frequently collide with the atoms of the conductor, as a result, they acquire a constant speed called the drift speed which is given by V_e = 1/enA where I = current in the conductor due to drifting electrons, e = charge of electron, n = number of free electrons per unit volume of the conductor and A = area of cross-section of the conductor. Choose the current statements

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