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NVAEasyJEE 2023Electric Current & Drift Velocity

JEE Physics 2023 Question with Solution

A current of 2 A2 \, \text{A} through a wire of cross-sectional area 25.0 mm225.0\,mm^2. The number of free electrons in a cubic meter is 2.0×10282.0 \times 10^{28}. The drift velocity of the electrons is _____ × 10−6 m/s10^{-6}\,m/s (given, charge on electron =1.6×10−19 C= 1.6 \times 10^{-19}\,C):

Answer

Correct answer:25

Step-by-step solution

Standard Method

Given: I=2 AI = 2 \, \text{A}, n=2×1028 m−3n = 2 \times 10^{28} \, \text{m}^{-3}, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C}, and cross-sectional area A=25×10−6 m2A = 25 \times 10^{-6} \, \text{m}^2.

Find: The drift velocity of electrons.

Use the formula for drift velocity:

Vd=IneAV_d = \frac{I}{n e A}

Substitute the given values:

Vd=2(2×1028)(1.6×10−19)(25×10−6)V_d = \frac{2}{(2 \times 10^{28})(1.6 \times 10^{-19})(25 \times 10^{-6})}

Simplifying:

Vd=2800×10−6=25×10−6 m/sV_d = \frac{2}{800 \times 10^{-6}} = 25 \times 10^{-6} \, \text{m/s}

Therefore, the drift velocity of the electrons is 25×10−6 m/s25 \times 10^{-6} \, \text{m/s}. Hence, the required numerical value is 25.

Step-by-step Explanation

Given: A very large number of free electrons are available per unit volume, so the drift velocity is expected to be very small.

Find: The numerical value multiplying 10−6 m/s10^{-6} \, \text{m/s}.

The relation between current and drift velocity is:

I=neAVdI = n e A V_d

So,

Vd=IneAV_d = \frac{I}{n e A}

Now substitute the values from the question:

Vd=2(2×1028)(1.6×10−19)(25×10−6)V_d = \frac{2}{(2 \times 10^{28})(1.6 \times 10^{-19})(25 \times 10^{-6})}

Evaluate the denominator as shown:

(2)(1.6)(25)=80(2)(1.6)(25) = 80

and the powers of ten combine to give:

1028×10−19×10−6=10310^{28} \times 10^{-19} \times 10^{-6} = 10^3

Hence the denominator becomes:

80×103=8×10480 \times 10^3 = 8 \times 10^4

Thus,

Vd=28×104=25×10−6 m/sV_d = \frac{2}{8 \times 10^4} = 25 \times 10^{-6} \, \text{m/s}

Therefore, the required answer is 25.

Common mistakes

  • Using the area in mm2mm^2 without converting it properly to SI units is incorrect. The formula requires area in m2\text{m}^2. Convert carefully before substitution.

  • Substituting the formula incorrectly as Vd=IneAV_d = I n e A is wrong. The drift velocity is inversely proportional to nn, ee, and AA, so use Vd=IneAV_d = \frac{I}{n e A}.

  • Losing track of powers of ten leads to an incorrect final magnitude. Combine the exponents systematically and keep the result in the form ×10−6\times 10^{-6} as asked.

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