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QJEEM24Ph2857: JEE Main 2024 Physics - Class 12 JEEM24Ph2857 Solution: The electric current through a wire varies with time as , where …

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

Multiple Choice Options

  • A 80 C
  • B 800 C
  • C 1000 C
  • D 1600 C

Step-by-Step Verified Solution

To calculate the amount of electric charge QQ that crosses through a section of the wire over a period of 2020 seconds, given the current varies with time as I=I0+βtI = I_0 + \beta t , where I0=20I_0 = 20 A (initial current) and β=3\beta = 3 A/s (rate of change of current with time), we use the concept of integration from calculus because the current is not constant but changes linearly with time.

The electric charge QQ is the integral of current II over the time interval from 00 to 2020 seconds. The formula for QQ is given by:

Q=∫0TI(t) dtQ = \int_{0}^{T} I(t) \, dt

Where I(t)=I0+βtI(t) = I_0 + \beta t and T=20T = 20 s. Substituting the given values:

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

This integral can be solved in two parts:

1. The integral of the constant term 2020 :

∫20 dt=20t\int 20 \, dt = 20t

2. The integral of the linear term 3t3t :

∫3t dt=32t2\int 3t \, dt = \frac{3}{2}t^2

Thus, combining these and evaluating from 00 to 2020 seconds:

Q=[20t+32t2]020Q = \left[20t + \frac{3}{2}t^2\right]_{0}^{20}

Q=[20(20)+32(20)2]−[20(0)+32(0)2]Q = \left[20(20) + \frac{3}{2}(20)^2\right] - \left[20(0) + \frac{3}{2}(0)^2\right]

Q=400+600=1000Q = 400 + 600 = 1000 C

Therefore, the amount of electric charge that crosses through a section of the wire in 2020 seconds is 10001000 C.

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