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):

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

Now, we can calculate the drift velocity:

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

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:

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

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:

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

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:

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

Now, we can calculate the current:

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

Finally, we can calculate the drift velocity:

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

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:

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

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:

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

Therefore, the resistance of the conductor is 4 Ω.