Typical Numerical Problems on Drift Velocity and Resistance

JEE Exam:

1. Potential difference: 10 V Length: 1 m Cross-sectional area: 1 mm² = 1 x 10^-6 m² Resistivity: 1 x 10^-6 Ωm

Solution:

Drift velocity (v_d) can be calculated using the formula:

$$v_d = \frac{E}{nAq\rho}$$

Where,

• E = Electric field strength (V/m)
• n = Number of mobile charge carriers per unit volume (m^-3)
• A = Cross-sectional area of the conductor (m²)
• q = Charge of an electron (1.6 x 10^-19 C)
• ρ = Resistivity of the conductor (Ωm)

First, we need to calculate the electric field strength (E):

$$E = \frac{V}{L} = \frac{10 V}{1 m} = 10 V/m$$

Now, we can calculate the drift velocity:

$$v_d = \frac{E}{nAq\rho} = \frac{10 V/m}{(8.5 x 10^{28} m^{-3})(1 x 10^{-6} m^2)(1.6 x 10^{-19} C)(1 x 10^{-6} \Omega m)}$$ $$v_d \approx 7.35 x 10^{-5} m/s$$

Therefore, the drift velocity of electrons in the conductor is approximately 7.35 x 10^-5 m/s.

2. Current: 1 A Length: 1 m Cross-sectional area: 1 mm² = 1 x 10^-6 m² Resistivity: 1 x 10^-6 Ωm

Solution:

In this case, we can directly use the formula for drift velocity:

$$v_d = \frac{I}{nAq}$$

Where,

• I = Current (A)
• n = Number of mobile charge carriers per unit volume (m^-3)
• A = Cross-sectional area of the conductor (m²)
• q = Charge of an electron (1.6 x 10^-19 C)

Substituting the given values:

$$v_d = \frac{1 A}{(8.5 x 10^{28} m^{-3})(1 x 10^{-6} m^2)(1.6 x 10^{-19} C)}$$ $$v_d \approx 7.35 x 10^{-5} m/s$$

Therefore, the drift velocity of electrons in the conductor is 7.35 x 10^-5 m/s, which is the same as in the previous case.

3. Potential difference: 1 V Length: 1 cm = 0.01 m Cross-sectional area: 1 cm² = 1 x 10^-4 m² Resistivity: 1 x 10^-4 Ωm

Solution:

To find the resistance (R) of the conductor, we can use the formula:

$$R = \rho \frac{L}{A}$$

Where,

• ρ = Resistivity of the conductor (Ωm)
• L = Length of the conductor (m)
• A = Cross-sectional area of the conductor (m²)

Substituting the given values:

$$R = (1 x 10^{-4} \Omega m) \frac{0.01 m}{1 x 10^{-4} m^2}$$ $$R = 1 \Omega$$

Therefore, the resistance of the conductor is 1 Ω.

CBSE Board Exam:

1. Potential difference: 6 V Length: 2 m Cross-sectional area: 2 mm² = 2 x 10^-6 m² Resistivity: 2 x 10^-6 Ωm

Solution:

To calculate the drift velocity, we’ll use the formula:

$$v_d = \frac{I}{nAq}$$

Where,

• I = Current (A)
• n = Number of mobile charge carriers per unit volume (m^-3)
• A = Cross-sectional area of the conductor (m²)
• q = Charge of an electron (1.6 x 10^-19 C)

First, we need to find the current (I) using Ohm’s law:

$$I = \frac{V}{R}$$

Since the resistance (R) is not given, we can use the formula:

$$R = \rho \frac{L}{A}$$

Substituting the given values:

$$R = (2 x 10^{-6} \Omega m) \frac{2 m}{2 x 10^{-6} m^2}$$ $$R = 2 \Omega$$

Now, we can calculate the current:

$$I = \frac{V}{R} = \frac{6 V}{2 \Omega} = 3 A$$

Finally, we can calculate the drift velocity:

$$v_d = \frac{I}{nAq}$$ $$v_d = \frac{3 A}{(8.5 x 10^{28} m^{-3})(2 x 10^{-6} m^2)(1.6 x 10^{-19} C)}$$ $$v_d \approx 1.10 x 10^{-4} m/s$$

Therefore, the drift velocity of electrons in the conductor is approximately 1.10 x 10^-4 m/s.

Note: The number of mobile charge carriers (n) is taken as the typical value for copper, which is 8.5 x 10²⁸ m^-3.

2. Current: 0.5 A Length: 3 m Cross-sectional area: 3 mm² = 3 x 10^-6 m² Resistivity: 3 x 10^-6 Ωm

Solution:

Using the formula for drift velocity:

$$v_d = \frac{I}{nAq}$$

Where,

• I = Current (A)
• n = Number of mobile charge carriers per unit volume (m^-3)
• A = Cross-sectional area of the conductor (m²)
• q = Charge of an electron (1.6 x 10^-19 C)

We can directly calculate the drift velocity:

$$v_d = \frac{0.5 A}{(8.5 x 10^{28} m^{-3})(3 x 10^{-6} m^2)(1.6 x 10^{-19} C)}$$ $$v_d \approx 1.10 x 10^{-4} m/s$$

Therefore, the drift velocity of electrons in the conductor is approximately 1.10 x 10^-4 m/s.

3. Potential difference: 2 V Length: 4 cm = 0.04 m Cross-sectional area: 4 cm² = 4 x 10^-4 m² Resistivity: 4 x 10^-4 Ωm

Solution:

To find the resistance (R) of the conductor, we can use the formula:

$$R = \rho \frac{L}{A}$$

Substituting the given values:

$$R = (4 x 10^{-4} \Omega m) \frac{0.04 m}{4 x 10^{-4} m^2}$$ $$R = 4 \Omega$$

Therefore, the resistance of the conductor is 4 Ω.