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MCQEasyJEE 2024Electric Current & Drift Velocity

JEE Physics 2024 Question with Solution

The electric current through a wire varies with time as I=I0+βtI = I_0 + \beta t, where I0=20 AI_0 = 20 \, \text{A} and β=3 A/s\beta = 3 \, \text{A/s}. The amount of electric charge that crosses through a section of the wire in 20 s20 \, \text{s} is:

  • A

    80 C80 \, \text{C}

  • B

    1000 C1000 \, \text{C}

  • C

    800 C800 \, \text{C}

  • D

    1600 C1600 \, \text{C}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: I=I0+βtI = I_0 + \beta t, where I0=20 AI_0 = 20 \, \text{A}, β=3 A/s\beta = 3 \, \text{A/s}, and time interval is from t=0t = 0 to t=20 st = 20 \, \text{s}.

Find: The total electric charge QQ crossing the wire in 20 s20 \, \text{s}.

Use the relation between current and charge:

Q=∫I dtQ = \int I \, dt

Substitute the given expression for current:

Q=∫020(I0+βt) dt=∫020(20+3t) dtQ = \int_0^{20} (I_0 + \beta t) \, dt = \int_0^{20} (20 + 3t) \, dt

Evaluate the integral:

Q=[20t+32t2]020Q = \left[20t + \frac{3}{2}t^2\right]_0^{20}

Now substitute the limits:

Q=(20×20+32×202)−(20×0+32×02)Q = \left(20 \times 20 + \frac{3}{2} \times 20^2\right) - \left(20 \times 0 + \frac{3}{2} \times 0^2\right)

Simplifying:

Q=400+32×400Q = 400 + \frac{3}{2} \times 400 Q=400+600Q = 400 + 600 Q=1000 CQ = 1000 \, \text{C}

Therefore, the amount of electric charge that crosses the wire is 1000 C1000 \, \text{C}. The correct option is B.

Integral Split Method

Given: I=I0+βt=20+3tI = I_0 + \beta t = 20 + 3t.

Find: Total charge qq from t=0t = 0 to t=20 st = 20 \, \text{s}.

Since current is the rate of flow of charge,

I=dqdtI = \frac{dq}{dt}

so,

dq=(20+3t) dtdq = (20 + 3t) \, dt

Integrate over the given time interval:

q=∫020(20+3t) dtq = \int_{0}^{20} (20 + 3t) \, dt

Split the integral:

q=∫02020 dt+∫0203t dtq = \int_{0}^{20} 20 \, dt + \int_{0}^{20} 3t \, dt

Evaluate each part:

q=[20t]020+[3t22]020q = \left[20t\right]_{0}^{20} + \left[\frac{3t^2}{2}\right]_{0}^{20}

Substitute the limits:

q=(20×20)+3×2022q = (20 \times 20) + \frac{3 \times 20^2}{2} q=400+3×4002q = 400 + \frac{3 \times 400}{2} q=400+600=1000 Cq = 400 + 600 = 1000 \, \text{C}

So, the total charge is 1000 C1000 \, \text{C}, hence the correct option is B.

Common mistakes

  • Using Q=ItQ = It with a constant current assumption is incorrect because the current varies with time as I=I0+βtI = I_0 + \beta t. Instead, integrate the time-dependent current: Q=∫I dtQ = \int I \, dt.

  • Forgetting the integration limits from 00 to 20 s20 \, \text{s} gives an incomplete expression for charge. Always evaluate the definite integral over the full time interval.

  • Integrating 3t3t incorrectly as 3t23t^2 instead of 32t2\frac{3}{2}t^2 leads to an overestimated charge. Apply the power rule carefully while integrating.

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